MITCHELL's THEOREM REVISITED Contents Part 1. Basic Side

MITCHELL’S THEOREM REVISITED

THOMAS GILTON AND JOHN KRUEGER

Abstract. Mitchell’s theorem on the approachability ideal states that it is consistent relative to a greatly Mahlo cardinal that there is no stationary subset of ω2 ∩ cof(ω1) in the approachability ideal I[ω2]. In this paper we give a new proof of Mitchell’s theorem, deriving it from an abstract framework of side condition methods.

Contents

Part 1. Basic side condition methods 4 1. Adequate sets 4 2. Analysis of remainder points 13 3. Strong genericity and cardinal preservation 21 4. Adding a club 25 5. S~-obedient side conditions 28 6. The approximation property and factorization 31

Part 2. Advanced side condition methods 32 7. Mitchell’s use of κ 32 8. Interaction of models past κ 40 9. Canonical models 47 10. Closure under canonical models 54 11. The main proxy lemma 60 12. The proxy construction 67 13. Amalgamation of side conditions 77

Part 3. Mitchell’s Theorem 86 14. The ground model 86 15. The forcing poset 87 16. The ﬁnal argument 91 References 96

Date: June 2015; revised November 2016. 2010 Mathematics Subject Classiﬁcation: Primary 03E35, 03E40; Secondary 03E05. Key words and phrases. Forcing, approachability ideal, side conditions, adequate sets. 1 2 THOMAS GILTON AND JOHN KRUEGER

Introduction

The approachability ideal I[λ+], for an uncountable cardinal λ, is defined as + + follows. For a given sequence ~a = hai : i < λ i of bounded subsets of λ , let S~a denote the set of limit ordinals α < λ+ for which there exists a set c ⊆ α, which is club in α with order type cf(α), such that for all β < α, there is i < α with c ∩ β = ai. Intuitively speaking, the set S~a carries a kind of weak square sequence, namely a sequence of clubs such that for each α in S~a, the club attached to α has its initial segments enumerated at stages prior to α. Define I[λ+] as the collection of sets S ⊆ λ+ for which there exists a sequence ~a as above and a club C ⊆ λ+ + + such that S ∩ C ⊆ S~a. In other words, I[λ ] is the ideal of subsets of λ which is generated modulo the club filter by sets of the form S~a. Let λ be a regular uncountable cardinal. Shelah [14] proved that the set λ+ ∩ cof(< λ) is in I[λ+]. Therefore the structure of I[λ+] is determined by which + ∗ subsets of λ ∩ cof(λ) belong to it. At one extreme, the weak square principle λ implies that λ+ ∩cof(λ) is in I[λ+]; therefore I[λ+] is just the power set of λ+. The opposite extreme would be that no stationary subset of λ+ ∩cof(λ) belongs to I[λ+], in other words, that I[λ+] is the nonstationary ideal when restricted to cofinality λ. Whether the second extreme is consistent was open for several decades, and was eventually solved by Mitchell [12]. Mitchell proved that it is consistent, relative to the consistency of a greatly Mahlo cardinal, that there does not exist a stationary subset of ω2 ∩ cof(ω1) in I[ω2]. We will refer to this result as Mitchell’s theorem. Mitchell’s theorem is important not only for solving a deep and long-standing open problem in combinatorial set theory, but also for introducing powerful new techniques in forcing. A basic tool in the proof is a forcing poset for adding a club subset of ω2 with finite conditions, using finite sets of countable models as side conditions. A similar forcing poset was introduced by Friedman [3] around the same time. The use of countable models in Friedman’s and Mitchell’s forcing posets for adding a club expanded the original side condition method of Todorˇcević[15], which was designed to add a generic object of size ω1, to adding a generic object of size ω2. In addition, Mitchell’s proof introduced the new concepts of strongly generic conditions and strongly proper forcing posets, which are closely related to the approximation property. Several years later, Neeman [13] developed a general framework of side conditions, which he called sequences of models of two types. An important distinction between Neeman’s side conditions and those of Friedman and Mitchell is that the two-type side conditions include both countable and uncountable models. A couple of years later, Krueger [6] developed an alternative framework of side conditions called adequate sets. This approach bases the analysis of side conditions on the ideas of the comparison point and remainder points of two countable models. No- tably, this approach has led to the solution of an open problem of Friedman [3], by showing how to add a club subset of ω2 with finite conditions while preserving the continuum hypothesis ([10]). Other applications are given in [8], [7], [9], and [2]. Notwithstanding the merits of the frameworks of Neeman [13] and Krueger [6], these frameworks are limited in the sense that they are intended to add a single subset of ω2 (or of a cardinal κ which is collapsed to become ω2). The proof of Mitchell’s theorem, on the other hand, involves adding κ+ many club subsets of a cardinal κ. Many consistency proofs in set theory about a cardinal κ involve adding MITCHELL’S THEOREM REVISITED 3

κ+ many subsets of κ by forcing, so that each of the potential counterexamples to the statement being forced is captured in some intermediate generic extension and dealt with by the rest of the forcing extension. The goal of this paper is to extend the framework of adequate sets to allow for adding many subsets of ω2, or of a cardinal κ which is collapsed to become ω2. The purpose of this extension is to provide general tools which will be useful for proving new consistency results on ω2. In Part III we give an example by deriving Mitchell’s theorem from the abstract framework developed in Parts I and II. The paper includes a very detailed treatment of adequate sets and remainder points in Sections 1 and 2, and of Mitchell’s application of the square principle to side conditions in Sections 7 and 8. We also develop some new ideas, including canonical models in Sections 9 and 10, and the main proxy lemma in Section 11. We will analyze ﬁnite sets of countable elementary substructures of H(κ+). The method of adequate sets handles the interaction of the models below κ. Following Mitchell, we employ the square principle κ to describe and control the interaction of countable models between κ and κ+. We introduce a new kind of side condition, which we call an S~-obedient side condition. We show that the forcing poset consisting of S~-obedient side conditions on H(κ+), where κ is a greatly Mahlo cardinal, ordered by component-wise inclusion, forces that κ = ω2 and there is no stationary subset of ω2 ∩ cof(ω1) in the approachability ideal I[ω2].

This project began with the M.S. thesis of Gilton at the University of North Texas, in which he reconstructed the original proof of Mitchell’s theorem in the context of adequate sets. Krueger is indebted to Gilton for explaining to him many of the details of Mitchell’s proof, especially the use of κ. Gilton isolated a workable requirement on remainder points which later evolved into the idea of S~-obedient side conditions. After Gilton’s thesis was complete, Krueger returned to the problem and made a number of advances. Krueger developed the new idea of canonical models, which is dealt with in Sections 9 and 10. Canonical models are models which appear in a given model N, reflect information about models lying outside of N, and are determined by canonical parameters which arise in the comparison of models. He isolated the main proxy lemma, Lemma 11.5, which significantly simplifies the method of proxies used by Mitchell. And he introduced the idea of S~-obedient side conditions, and showed that forcing with pure side conditions on a greatly Mahlo cardinal produces a generic extension in which the approachability ideal on ω2 restricted to cofinality ω1 is the nonstationary ideal.

This paper was written for an audience with a minimum background of one year of graduate studies in set theory, with a working knowledge of forcing and proper forcing, and with some familiarity with generalized stationarity. For a regular uncountable cardinal µ and a set X with µ ⊆ X, we let Pµ(X) denote the set {a ⊆ X : |a| < µ}. A set S ⊆ Pµ(X) being stationary is equivalent to the statement that for any function F : X<ω → X, there exists a ∈ S such that a ∩ µ ∈ µ and a is closed under F . If a is a set of ordinals, then lim(a) denotes the set of ordinals β such that for all γ < β, a ∩ (γ, β) 6= ∅. We let cl(a) = a ∪ lim(a). If M is a set, we write sup(M) to denote sup(M ∩ On). 4 THOMAS GILTON AND JOHN KRUEGER

If A is a structure in a ﬁrst order language, and X1,...,Xk are subsets of the underlying set of A, then we write (A,X1,...,Xk) to denote the expansion of the structure A obtained by adding X1,...,Xk as predicates.

Part 1. Basic side condition methods

§1. Adequate sets

We begin the paper by working out the basic framework of adequate sets. Roughly speaking, this framework provides methods for describing and handling the interaction of countable elementary substructures below ω2, or below κ for some regular uncountable cardinal κ which is intended to become ω2 in a forcing extension. Adequate sets were introduced by Krueger [6]; many of the results of this section appear in [6], although in a slightly diﬀerent form.

∗ ∗ ∗ We ﬁx objects κ, λ, T , π , C , Λ, X0, and Y0 as follows.

Notation 1.1. For the remainder of the paper, κ is a regular cardinal with ω2 ≤ κ.

In [6] we only considered the case when κ = ω2. In the proof of Mitchell’s theorem given in Part III, κ is a greatly Mahlo cardinal. Notation 1.2. Fix a cardinal λ such that κ ≤ λ. In Parts II and III we will let λ = κ+.

Deﬁnition 1.3. A set T ⊆ Pω1 (κ) is thin if for all β < κ, |{a ∩ β : a ∈ T }| < κ. The idea of a thin stationary set was introduced by Friedman [3], who used a thin stationary set to develop a forcing poset for adding a club subset of a fat stationary subset of ω2 with ﬁnite conditions. ω Observe that if |β | < κ for all β < κ, then Pω1 (κ) itself is thin. Krueger proved that the existence of a thin stationary subset of Pω1 (ω2) is independent of ZFC; see [4].

∗ Notation 1.4. Fix a thin stationary set T ⊆ Pω1 (κ) which satisﬁes the property ∗ ∗ ∗ that for all β < κ and a ∈ T , a ∩ β ∈ T . In Part III, we will let T = Pω1 (κ). Note that if T is a thin stationary set, then the set {a ∩ β : a ∈ T, β < κ} is a thin stationary set which satisﬁes the property of being closed under initial segments which is described in Notation 1.4. Observe that if T is a thin stationary set, then |T | = κ. Notation 1.5. Fix a bijection π∗ : T ∗ → κ. Notation 1.6. Let C∗ denote the set of β < κ such that whenever a is a bounded subset of β in T ∗, then π∗(a) < β. The fact that T ∗ is thin easily implies that C∗ is a club subset of κ. Notation 1.7. Let Λ denote the set C∗ ∩ cof(>ω). MITCHELL’S THEOREM REVISITED 5

Notation 1.8. For the remainder of the paper, let E denote a well-ordering of H(λ). ∗ Notation 1.9. Let X0 denote the set of M in Pω1 (H(λ)) such that M ∩ κ ∈ T ∗ ∗ ∗ and M is an elementary substructure of (H(λ), ∈, E, κ, T , π ,C , Λ).

Notation 1.10. Let Y0 denote the set of P in Pκ(H(λ)) such that P ∩ κ ∈ κ and ∗ ∗ ∗ P is an elementary substructure of (H(λ), ∈, E, κ, T , π ,C , Λ).

Note that if P and Q are in Y0, then P ∩Q is in Y0. And if M ∈ X0 and P ∈ Y0, then M ∩ P is in X0. For the presence of the well-ordering E implies that P ∩ Q and M ∩ P are elementary substructures, and M ∩ P ∩ κ is an initial segment of ∗ M ∩ κ and hence is in T . For the intersection of models in X0, see Lemma 1.23. This completes the introduction of the basic objects.

Next we will deﬁne comparison points and a way to compare two models in X0.

Deﬁnition 1.11. For M ∈ X0, let ΛM denote the set of β ∈ Λ such that β = min(Λ \ sup(M ∩ β)).

Observe that since any member of ΛM is determined by an ordinal in cl(M), and cl(M) is countable, it follows that ΛM is countable.

Lemma 1.12. Let M ∈ X0. If β ∈ ΛM and β0 ∈ Λ ∩ β, then M ∩ [β0, β) 6= ∅.

Proof. If M ∩ [β0, β) = ∅, then sup(M ∩ β) ≤ β0. So

β = min(Λ \ sup(M ∩ β)) ≤ β0 < β, which is a contradiction.

Lemma 1.13. Let M and N be in X0. Then ΛM ∩ ΛN has a maximum element.

Proof. Note that the first member of Λ is in both ΛM and ΛN , and therefore ΛM ∩ ΛN is nonempty. Suppose for a contradiction that γ := sup(ΛM ∩ ΛN ) is not in ΛM ∩ ΛN . Fix an increasing sequence hγn : n < ωi in ΛM ∩ ΛN which is cofinal in γ. Then for each n < ω, M ∩ [γn, γn+1) is nonempty by Lemma 1.12. So γ is a limit point of M. Similarly, γ is a limit point of N. Let β = min(Λ \ γ). Since γ has cofinality ω, γ < β, and since γ is a limit point of M and a limit point of N, easily β ∈ ΛM ∩ ΛN . This contradicts that γ = sup(ΛM ∩ ΛN ) and γ < β.

Deﬁnition 1.14. For M and N in X0, let βM,N be the maximum element of ΛM ∩ ΛN . The ordinal βM,N is called the comparison point of M and N.

The most important property of βM,N is described in the next lemma.

Lemma 1.15. Let M and N be in X0. Then

cl(M ∩ κ) ∩ cl(N ∩ κ) ⊆ βM,N .

Proof. Suppose for a contradiction that ξ is in cl(M ∩ κ) ∩ cl(N ∩ κ) but βM,N ≤ ξ. Let β = min(Λ \ (ξ + 1)). Since β is a limit ordinal, βM,N ≤ ξ < ξ + 1 < β. We claim that β ∈ ΛM ∩ ΛN . Then by the maximality of βM,N , β ≤ βM,N , which is a contradiction. First, assume that ξ ∈ Λ. Then ξ has uncountable coﬁnality. So ξ cannot be a limit point of M or of N. Hence ξ ∈ M ∩ N. By elementarity, ξ + 1 ∈ M ∩ N. Since ξ + 1 ∈ M ∩ β, ξ + 1 ≤ sup(M ∩ β) < β. As β = min(Λ \ (ξ + 1)), clearly 6 THOMAS GILTON AND JOHN KRUEGER

β = min(Λ \ sup(M ∩ β)). So β ∈ ΛM . The same argument shows that β ∈ ΛN , and we are done. Secondly, assume that ξ∈ / Λ. Then min(Λ\ξ) = min(Λ\(ξ+1)) = β. Since ξ < β and ξ is either in M ∩ κ or is a limit point of M ∩ κ, clearly ξ ≤ sup(M ∩ β) < β. Hence β = min(Λ \ sup(M ∩ β)), and therefore β ∈ ΛM . The same argument shows that β ∈ ΛN , ﬁnishing the proof. The next lemma provides some useful technical facts about comparison points. Statement (4) is not very intuitive; however it turns out that this observation simpliﬁes some of the material in the original development of adequate sets in [6].

Lemma 1.16. Let L, M, and N be in X0.

(1) If L ∩ κ ⊆ M ∩ κ then ΛL ⊆ ΛM . Hence βL,N ≤ βM,N . (2) If L ∩ κ ⊆ β where β ∈ Λ, then ΛL ⊆ β + 1. Hence βL,M ≤ β. (3) If β < βM,N and β ∈ Λ, then M ∩ [β, βM,N ) 6= ∅. (4) Suppose that M ∩ βL,M ⊆ N. Then βL,M ≤ βL,N . Proof. Statements (1) and (2) can be proven in a straightforward way from the definitions, and (3) follows immediately from Lemma 1.12. (4) By definition, βL,M ∈ ΛL. Since M ∩ βL,M ⊆ N, sup(M ∩ βL,M ) ≤ sup(N ∩ βL,M ). As βL,M ∈ ΛM , by definition βL,M = min(Λ \ sup(M ∩ βL,M )). So clearly βL,M = min(Λ \ sup(N ∩ βL,M )). Hence βL,M ∈ ΛN . So βL,M ∈ ΛL ∩ ΛN . Therefore βL,M ≤ max(ΛL ∩ ΛN ) = βL,N .

Now we introduce our way of comparing models.

Deﬁnition 1.17. Let M and N be in X0.

(1) Let M < N if M ∩ βM,N ∈ N. (2) Let M ∼ N if M ∩ βM,N = N ∩ βM,N . (3) Let M ≤ N if either M < N or M ∼ N.

Deﬁnition 1.18. A ﬁnite set A ⊆ X0 is said to be adequate if for all M and N in A, either M < N, M ∼ N, or N < M.

If M < N, then by elementarity cl(M ∩ βM,N ) is a member of N. Since cl(M ∩ βM,N ) is countable, cl(M ∩ βM,N ) ⊆ N. Also every initial segment of M ∩ βM,N is in N. For any proper initial segment has the form M ∩ γ = M ∩ βM,N ∩ γ for some γ ∈ M ∩ βM,N , and since M ∩ βM,N and γ are in N, so is M ∩ γ. The next lemma provides some useful technical facts about the relation on models just introduced. Lemma 1.19. Let {M,N} be adequate.

(1) If (N ∩ βM,N ) \ M is nonempty, then M < N. (2) If M ≤ N then M ∩ βM,N = M ∩ N ∩ κ = M ∩ N ∩ βM,N . (3) βM,N = min(Λ \ sup(M ∩ N ∩ κ)). (4) If M < N then βM,N ∈ N. (5) If β < βM,N and β ∈ Λ, then (M ∩ N) ∩ [β, βM,N ) 6= ∅. Proof. The assumption of (1) implies that M ∼ N and N < M are impossible. (2) Both M ∩ βM,N ∈ N and M ∩ βM,N = N ∩ βM,N imply that M ∩ βM,N ⊆ N. So M ∩ βM,N ⊆ M ∩ N ∩ κ. Conversely by Lemma 1.15, M ∩ N ∩ κ ⊆ βM,N , MITCHELL’S THEOREM REVISITED 7 so M ∩ N ∩ κ ⊆ M ∩ βM,N . This proves that M ∩ N ∩ κ = M ∩ βM,N . Since M ∩ N ∩ κ ⊆ βM,N by Lemma 1.15, M ∩ N ∩ κ = M ∩ N ∩ βM,N . (3) Without loss of generality assume that M ≤ N. Then M ∩N ∩κ = M ∩βM,N by (2). Since βM,N ∈ ΛM , by deﬁnition

βM,N = min(Λ \ (sup(M ∩ βM,N ))) = min(Λ \ (sup(M ∩ N ∩ κ))).

(4) If M < N then M ∩ βM,N ∈ N. By (2), M ∩ βM,N = M ∩ N ∩ κ. So M ∩ N ∩ κ ∈ N. By (3), βM,N = min(Λ \ sup(M ∩ N ∩ κ)). So βM,N ∈ N by elementarity. (5) Without loss of generality assume that M ≤ N. Then by (2), M ∩N ∩βM,N = M ∩βM,N . Since βM,N ∈ ΛM , Lemma 1.12 implies that M ∩[β, βM,N ) is nonempty. Fix ξ ∈ M ∩[β, βM,N ). Then ξ ∈ M ∩βM,N = M ∩N ∩βM,N . So (M ∩N)∩[β, βM,N ) is nonempty.

Lemma 1.20. Let M and N be in X0, and assume that {M,N} is adequate. Then cl(M ∩ N ∩ κ) = cl(M ∩ κ) ∩ cl(N ∩ κ). Proof. The forward inclusion is immediate. Suppose that α is in cl(M∩κ)∩cl(N∩κ). Then by Lemma 1.15, α < βM,N . Without loss of generality, assume that M ≤ N. Then α ∈ cl(M ∩ κ) ∩ βM,N = cl(M ∩ βM,N ) = cl(M ∩ N ∩ κ) by Lemma 1.19(2). If {M,N} is adequate, then the relation which holds between M and N is determined by the intersection of M and N with ω1. Lemma 1.21. Let {M,N} be adequate. Then:

(1) M < N iﬀ M ∩ ω1 < N ∩ ω1; (2) M ∼ N iﬀ M ∩ ω1 = N ∩ ω1.

Proof. Suppose that M < N. Then M ∩ βM,N ∈ N. Since βM,N has uncountable coﬁnality, ω1 ≤ βM,N . So M ∩ ω1 is an initial segment of M ∩ βM,N , and hence M ∩ ω1 ∈ N. So M ∩ ω1 < N ∩ ω1. Suppose that M ∼ N. Then M ∩ βM,N = N ∩ βM,N . Since ω1 ≤ βM,N , M ∩ ω1 = N ∩ ω1. Conversely if M ∩ ω1 < N ∩ ω1, then the facts just proved imply that M < N is the only possibility of how M and N relate. Similarly M ∩ ω1 = N ∩ ω1 implies that M ∼ N. Lemma 1.22. Let A be an adequate set. Then the relation < is irreﬂexive and transitive on A, ∼ is an equivalence relation on A, and the relations < and ≤ respect ∼.

Proof. Immediate from Lemma 1.21.

In proving amalgamation results over countable models, we will need to be able to enlarge an adequate set A by adding M ∩ N to A, where M < N are in A. Let us show that we can do this while preserving adequacy. First we note that M ∩ N is in X0.

Lemma 1.23. Let {M,N} be adequate. Then M ∩ N is in X0. 8 THOMAS GILTON AND JOHN KRUEGER

Proof. Without loss of generality, assume that M ≤ N. Then by Lemma 1.19(2), ∗ ∗ M ∩ N ∩ κ = M ∩ βM,N . Since T is closed under initial segments and M ∩ κ ∈ T , ∗ ∗ it follows that M ∩ βM,N ∈ T . Hence M ∩ N ∩ κ ∈ T . Also clearly M ∩ N is an elementary substructure.

Lemma 1.24. Let K, M, and N be in X0. Suppose that M < N and {K,M} is adequate. Then:

(1) βK,M∩N ≤ βK,M and βK,M∩N ≤ βM,N ; (2) M < K iff M ∩ N < K; (3) K ∼ M iff K ∼ M ∩ N; (4) K < M iff K < M ∩ N. In particular, {K,M ∩ N} is adequate.

Proof. (1) Since M∩N ⊆ M, βK,M∩N ≤ βK,M by Lemma 1.16(1). Also M∩N∩κ ⊆ βM,N by Lemma 1.15, which implies that βK,M∩N ≤ βM,N by Lemma 1.16(2). This proves (1). Since M ∩ N ∩ βM,N = M ∩ βM,N by Lemma 1.19(2) and βK,M∩N ≤ βM,N , it follows that

M ∩ N ∩ βK,M∩N = M ∩ βK,M∩N . (2,3,4) First we will prove the forward implications of (2), (3), and (4). If M < K then M ∩ βK,M is in K. But since βK,M∩N ≤ βK,M , M ∩ βK,M∩N is an initial segment of M ∩ βK,M , and hence is in K. So M ∩ N ∩ βK,M∩N = M ∩ βK,M∩N is in K, and therefore M ∩ N < K. If K ∼ M, then K ∩ βK,M = M ∩ βK,M . Since βK,M∩N ≤ βK,M ,

K ∩ βK,M∩N = M ∩ βK,M∩N = M ∩ N ∩ βK,M∩N .

Therefore K ∼ M ∩ N. Suppose that K < M. Then K ∩ βK,M ∈ M. Since βK,M∩N ≤ βK,M , K ∩ βK,M∩N ∈ M. So to show that K < M ∩ N, it suffices to show that K ∩ βK,M∩N ∈ N. ∗ ∗ ∗ Since K ∩κ ∈ T by the definition of X0, K ∩βK,M∩N ∈ T as T is closed under initial segments. Recall from Notation 1.5 that π∗ : T ∗ → κ is a bijection. As M is ∗ ∗ closed under π by elementarity, π (K ∩ βK,M∩N ) ∈ M ∩ κ. Since K ∩ βK,M∩N is a bounded subset of βK,M∩N and βK,M∩N ≤ βM,N , we have that K ∩ βK,M∩N is a ∗ bounded subset of βM,N . Since βM,N ∈ Λ, it follows that π (K ∩ βK,M∩N ) < βM,N ∗ ∗ by the definitions of C and Λ from Notations 1.6 and 1.7. Hence π (K∩βK,M∩N ) ∈ ∗ M ∩ βM,N ⊆ N. Since N is closed under the inverse of π by elementarity, K ∩ βK,M∩N ∈ N. Now we consider the reverse implications of (2), (3), and (4). Suppose that M ∩ N < K. Since {K,M} is adequate, either K < M, K ∼ M, or M < K. But K ∼ M and K < M are ruled out by the forward implications of (3) and (4). So M < K. The other converses are proved similarly.

Proposition 1.25. Let A be an adequate set and N ∈ X0. Let M be in A, and suppose that M < N. Then A ∪ {M ∩ N} is adequate.

Proof. Immediate from Lemma 1.24. MITCHELL’S THEOREM REVISITED 9

Our next goal is to prove the ﬁrst amalgamation result over countable models, which is stated in Proposition 1.29 below. See Proposition 13.1 for a much deeper result.

Lemma 1.26. Let L, M, and N be in X0. Suppose that N ≤ M and L ∈ N. Then L < M.

∗ Proof. Since L ∈ N, βL,M ≤ βM,N by Lemma 1.16(1). Also L ∩ βL,M is in N ∩ T , since it is an initial segment of L ∩ κ. As N is closed under π∗ by elementarity, the ∗ ordinal π (L ∩ βL,M ) is in N ∩ κ. And as βM,N ∈ Λ and L ∩ βL,M is a bounded ∗ ∗ subset of βM,N in T , it follows that π (L ∩ βL,M ) < βM,N by the deﬁnition of Λ. ∗ Hence π (L ∩ βL,M ) ∈ N ∩ βM,N ⊆ M. By elementarity, M is closed under the ∗ inverse of π , so L ∩ βL,M ∈ M.

Lemma 1.27. Let L, M, and N be in X0. Suppose that M < N and L ∈ N. Then:

(1) βL,M = βL,M∩N ; (2) L ∼ M ∩ N iff L ∼ M; (3) L < M ∩ N iff L < M; (4) M ∩ N < L iff M < L.

Proof. (1) Since M ∩ N ∩ κ ⊆ M ∩ κ, βL,M∩N ≤ βL,M by Lemma 1.16(1), which proves one direction of the equality. Since L ∩ κ ⊆ N ∩ κ, βL,M ≤ βM,N by Lemma 1.16(1). So M ∩ βL,M ⊆ M ∩ βM,N ⊆ M ∩ N.

By Lemma 1.16(4), βL,M ≤ βL,M∩N . (2,3,4) First we will prove the forward implications of (2), (3), and (4). As βL,M ≤ βM,N and M ∩ βM,N = M ∩ N ∩ βM,N , it follows that

M ∩ βL,M = M ∩ N ∩ βL,M . If L ∼ M ∩ N, then

L ∩ βL,M = L ∩ βL,M∩N = M ∩ N ∩ βL,M∩N = M ∩ N ∩ βL,M = M ∩ βL,M .

So L ∩ βL,M = M ∩ βL,M , and hence L ∼ M. And if L < M ∩ N, then

L ∩ βL,M = L ∩ βL,M∩N ∈ M ∩ N ⊆ M.

So L ∩ βL,M ∈ M, and hence L < M. If M ∩ N < L, then

M ∩ βL,M = M ∩ N ∩ βL,M = M ∩ N ∩ βL,M∩N ∈ L.

So M ∩ βL,M ∈ L, and therefore M < L. For the reverse implications, each of the assumptions L ∼ M, L < M, and M < L implies that {L, M} is adequate. Hence these assumptions imply that L ∼ M ∩ N, L < M ∩ N, and M ∩ N < L respectively by Lemma 1.24.

Lemma 1.28. Let L, M, and N be in X0. Suppose that M < N and L ∈ N. If {L, M ∩ N} is adequate, then {L, M} is adequate.

Proof. Immediate from Lemma 1.27. Proposition 1.29. Let A be adequate, N ∈ A, and suppose that for all M ∈ A, if M < N then M ∩ N ∈ A ∩ N. Suppose that B is adequate and A ∩ N ⊆ B ⊆ N. Then A ∪ B is adequate. 10 THOMAS GILTON AND JOHN KRUEGER

Proof. Let L ∈ B and M ∈ A, and we will show that {L, M} is adequate. Since L ∈ B and B ⊆ N, L ∈ N. If N ≤ M, then L < M by Lemma 1.26. Suppose that M < N. Then M ∩ N ∈ A ∩ N by assumption. Since A ∩ N ⊆ B, M ∩ N ∈ B. As B is adequate, {L, M ∩ N} is adequate. Since L ∈ N and {L, M ∩ N} is adequate, {L, M} is adequate by Lemma 1.28. In the last proposition, we assumed that M < N implies that M ∩ N ∈ N, for M ∈ A. At this point we do not have any reason to believe this implication is true in general. In Section 7, we will deﬁne a subclass of X0 on which this implication holds. See Notation 7.7 and Lemma 8.2.

So far we have discussed the interaction of countable models in X0. We now turn our attention to how models in X0 relate to models in Y0.

Lemma 1.30. Let M and N be in X0 ∪ Y0. Suppose that:

(1) M and N are in X0 and M < N, or (2) M and N are in Y0 and M ∩ κ < N ∩ κ, or (3) M ∈ X0, N ∈ Y0, and sup(M ∩ N ∩ κ) < N ∩ κ. Then M ∩ N ∩ κ ∈ N.

Proof. (1) If M and N are in X0, then M < N implies that M ∩ βM,N ∈ N. By Lemma 1.19(2), M ∩ βM,N = M ∩ N ∩ κ, so M ∩ N ∩ κ ∈ N. (2) If M and N are in Y0, then since M ∩ κ < N ∩ κ, M ∩ N ∩ κ = M ∩ κ ∈ N. (3) Suppose that M ∈ X0, N ∈ Y0, and sup(M ∩N ∩κ) < N ∩κ. Let β := N ∩κ. By the elementarity of N, β is a limit point of Λ. So fix γ ∈ N ∩ Λ such that sup(M ∩ β) < γ. Then M ∩ N ∩ κ = M ∩ γ. Since γ has uncountable cofinality, ∗ M ∩ γ is a bounded subset of γ, and as M ∈ X0, M ∩ γ ∈ T . By the definition of C∗ and Λ, π∗(M ∩ γ) < γ < N ∩ κ. Since N is closed under the inverse of π∗ by elementarity, M ∩ γ = M ∩ N ∩ κ ∈ N. Note that (3) holds if cf(N ∩ κ) > ω, which is the typical situation that we will consider.

Lemma 1.31. Let M ∈ X0 and N ∈ Y0, and assume that sup(M ∩N ∩κ) < N ∩κ. Then cl(M ∩ N ∩ κ) = cl(M ∩ κ) ∩ cl(N ∩ κ) ∩ (N ∩ κ). Proof. The forward inclusion is immediate. Let α ∈ cl(M ∩ κ) ∩ cl(N ∩ κ) ∩ (N ∩ κ). Since cl(N ∩ κ) = (N ∩ κ) ∪ {N ∩ κ}, α ∈ cl(M ∩ κ) ∩ (N ∩ κ) = cl(M ∩ N ∩ κ).

Recall that if M ∈ X0 and P ∈ Y0, then M ∩ P ∈ X0. We show next that we can add M ∩ P to an adequate set and preserve adequacy.

Lemma 1.32. Let K and M be in X0 and P in Y0. Assume that {K,M} is adequate and sup(M ∩ P ∩ κ) < P ∩ κ. Then:

(1) βK,M∩P ≤ βK,M and βK,M∩P < P ∩ κ; (2) M < K iﬀ M ∩ P < K; MITCHELL’S THEOREM REVISITED 11

(3) K ∼ M iﬀ K ∼ M ∩ P ; (4) K < M iﬀ K < M ∩ P . In particular, {K,M ∩ P } is adequate.

Proof. (1) Since M ∩ P ⊆ M, βK,M∩P ≤ βK,M by Lemma 1.16(1). As sup(M ∩ P ∩ κ) < P ∩ κ and Λ is unbounded in P ∩ κ by elementarity, we can ﬁx β ∈ Λ with sup(M ∩ P ∩ κ) < β < P ∩ κ. By Lemma 1.16(2),

βK,M∩P ≤ β < P ∩ κ. This proves (1). It follows that

M ∩ βK,M∩P = M ∩ P ∩ βK,M∩P . (2,3,4) First we will prove the forward implications of (2), (3), and (4). Assume that M < K. Then M ∩ βK,M ∈ K. Since βK,M∩P ≤ βK,M , M ∩ βK,M∩P ∈ K. So

M ∩ P ∩ βK,M∩P = M ∩ βK,M∩P ∈ K. Hence M ∩ P < K. Suppose that K ∼ M. Then K ∩ βK,M = M ∩ βK,M . Since βK,M∩P ≤ βK,M , it follows that

K ∩ βK,M∩P = M ∩ βK,M∩P = M ∩ P ∩ βK,M∩P . Therefore K ∼ M ∩ P . Finally, assume that K < M. Then K ∩ βK,M ∈ M. Since βK,M∩P ≤ βK,M , K ∩ βK,M∩P ∈ M. As βK,M∩P < P ∩ κ, by elementarity there is γ ∈ P ∩ Λ ∗ with βK,M∩P < γ. Then K ∩ βK,M∩P is a bounded subset of γ in T . Hence ∗ ∗ π (K ∩ βK,M∩P ) < γ. In particular, π (K ∩ βK,M∩P ) ∈ P ∩ κ. By the elementarity ∗ of P , P is closed under the inverse of π . So K ∩βK,M∩P ∈ P . Thus K ∩βK,M∩P ∈ M ∩ P , and therefore K < M ∩ P . Conversely, assume that M ∩ P < K. Since {K,M} is adequate, either M < K, M ∼ K, or K < M. But the forward implications of (3) and (4) rule out M ∼ K and K < M. Hence M < K. The other converses are proved similarly.

Proposition 1.33. Let A be an adequate set. Let M be in A and P in Y0, and assume that sup(M ∩ P ∩ κ) < P ∩ κ. Then A ∪ {M ∩ P } is adequate.

Proof. Immediate from Lemma 1.32.

Next we will prove an amalgamation result for uncountable models. See Propo- sition 13.2 for a deeper result.

Lemma 1.34. Let L and M be in X0 and P ∈ Y0. Assume that L ∈ P and sup(M ∩ P ∩ κ) < P ∩ κ. Then:

(1) βL,M = βL,M∩P and βL,M < P ∩ κ; (2) L ∼ M ∩ P iff L ∼ M; (3) M ∩ P < L iff M < L; (4) L < M ∩ P iff L < M. In particular, {L, M ∩ P } is adequate iff {L, M} is adequate. 12 THOMAS GILTON AND JOHN KRUEGER

Proof. (1) Since M ∩ P ⊆ M, βL,M∩P ≤ βL,M by Lemma 1.16(1), which proves one direction of the equality. As L ∈ P , by elementarity, ΛL ∈ P . Since ΛL is countable, ΛL ⊆ P . As βL,M ∈ ΛL, βL,M ∈ P ∩ κ. So M ∩ βL,M ⊆ M ∩ P . By Lemma 1.16(4), it follows that βL,M ≤ βL,M∩P . (2,3,4) First we will prove the forward implications of (2), (3), and (4). Since βL,M ∈ P as noted above,

M ∩ βL,M = M ∩ P ∩ βL,M . If L ∼ M ∩ P , then

L ∩ βL,M = L ∩ βL,M∩P = M ∩ P ∩ βL,M∩P = M ∩ P ∩ βL,M = M ∩ βL,M .

So L ∩ βL,M = M ∩ βL,M , and hence L ∼ M. If M ∩ P < L, then

M ∩ βL,M = M ∩ P ∩ βL,M = M ∩ P ∩ βL,M∩P ∈ L.

So M ∩ βL,M ∈ L, and therefore M < L. And if L < M ∩ P , then

L ∩ βL,M = L ∩ βL,M∩P ∈ M ∩ P ⊆ M.

So L ∩ βL,M ∈ M, and therefore L < M. Conversely, the assumptions M < L, L ∼ M, and L < M imply that {L, M} is adequate. Hence each of these assumptions imply that M ∩ P < L, L ∼ M ∩ P , L < M ∩ P respectively by Lemma 1.32.

Proposition 1.35. Let A be adequate, P ∈ Y0, and assume that for all M ∈ A, M ∩ P ∈ A ∩ P . Suppose that B is adequate and A ∩ P ⊆ B ⊆ P . Then A ∪ B is adequate. Proof. Let L ∈ B and M ∈ A. Then M ∩ P ∈ A ∩ P ⊆ B. Since B is adequate, {L, M ∩ P } is adequate. As M ∩ P ∈ P , sup(M ∩ P ∩ κ) < P ∩ κ. By Lemma 1.34, {L, M} is adequate.

We conclude the discussion about models in X0 and Y0 with the following useful lemma.

Lemma 1.36. Let M and N be in X0, and assume that {M,N} is adequate. Let P ∈ Y0. Then either βM,N = βM∩P,N , or P ∩ κ < βM,N .

Proof. Since M ∩ P ⊆ M, βM∩P,N ≤ βM,N . If βM,N = βM∩P,N , then we are done. So assume that βM∩P,N < βM,N . We claim that P ∩ κ < βM,N . Suppose for a contradiction that βM,N ≤ P ∩ κ. Since βM∩P,N < βM,N , by Lemma 1.19(5), we can ﬁx ξ ∈ (M ∩ N) ∩ [βM∩P,N , βM,N ).

As βM,N ≤ P ∩ κ, ξ ∈ (M ∩ N) ∩ P ∩ κ = (M ∩ P ) ∩ N ∩ κ.

Therefore ξ < βM∩P,N by Lemma 1.15, which contradicts the choice of ξ.

Finally, we prove an amalgamation result over transitive models. 0 0 0 Lemma 1.37. Let M, M , N, and N be in X0. Assume that M ∩ κ = M ∩ κ and N ∩ κ = N 0 ∩ κ. Then:

(1) βM,N = βM 0,N 0 ; MITCHELL’S THEOREM REVISITED 13

(2) M ∼ N iff M 0 ∼ N 0; (3) M < N iff M 0 < N 0; (4) N < M iff N 0 < M 0. In particular, {M,N} is adequate iff {M 0,N 0} is adequate.

0 0 Proof. (1) Since M ∩ κ ⊆ M ∩ κ and N ∩ κ ⊆ N ∩ κ, it follows that βM,N ≤ βM 0,N ≤ βM 0,N 0 by Lemma 1.16(1). Similarly, the reverse inclusions imply that βM 0,N 0 ≤ βM,N . So βM,N = βM 0,N 0 . (2,3,4) It suﬃces to prove the forward direction of the iﬀ’s of (2), (3), and (4), since the converses hold by symmetry. If M ∼ N, then 0 0 M ∩ βM 0,N 0 = M ∩ βM,N = N ∩ βM,N = N ∩ βM 0,N 0 , which proves (2). Suppose that M < N. Then 0 M ∩ βM 0,N 0 = M ∩ βM,N ∈ N. By elementarity, ∗ 0 0 π (M ∩ βM 0,N 0 ) ∈ N ∩ κ = N ∩ κ. 0 ∗ 0 0 Since N is closed under the inverse of π by elementarity, M ∩ βM 0,N 0 ∈ N . (4) is similar. Proposition 1.38. Let A be an adequate set. Assume that X ≺ (H(κ+), ∈), |X| = κ, and X ∩ κ+ ∈ κ+. Let B be an adequate set such that A ∩ X ⊆ B ⊆ X. Suppose that for all M ∈ A, there is M 0 ∈ B such that M ∩ κ = M 0 ∩ κ. Then A ∪ B is adequate. Proof. Let M ∈ A and K ∈ B be given. Fix M 0 ∈ B such that M ∩ κ = M 0 ∩ κ. As {M 0,K} ⊆ B, {M 0,K} is adequate. Therefore {M,K} is adequate by Lemma 1.37.

§2. Analysis of remainder points

In this section we will provide a detailed analysis of remainder points; some of these arguments appeared previously in [8] and [9], although in a less complete form. This analysis will be the foundation from which we derive the amalgamation results of Section 13.

Deﬁnition 2.1. Let {M,N} be adequate. Let RM (N), the set of remainder points of N over M, be deﬁned as the set of ζ satisfying either:

(1) ζ = min((N ∩ κ) \ βM,N ), provided that M ∼ N, or (2) there is γ ∈ (M ∩ κ) \ βM,N such that ζ = min((N ∩ κ) \ γ).

Note that if N < M, then βM,N ∈ M by Lemma 1.19(4). It follows that min((N ∩ κ) \ βM,N ) ∈ RM (N) by Deﬁnition 2.1(2). The next lemma describes some basic properties of remainder points. Lemma 2.2. Let {M,N} be adequate. Then:

(1) RM (N) ∩ cl(M ∩ κ) = ∅; (2) RM (N) is ﬁnite; (3) suppose that ζ ∈ RM (N) and ζ > min(RM (N) ∪ RN (M)); then σ := min((M ∩ κ) \ sup(N ∩ ζ)) ∈ RN (M) and ζ = min((N ∩ κ) \ σ). 14 THOMAS GILTON AND JOHN KRUEGER

Proof. (1) If ζ ∈ RM (N), then by deﬁnition, ζ ∈ N and βM,N ≤ ζ. Hence ζ∈ / cl(M ∩ κ) by Lemma 1.15. (2) Suppose for a contradiction that hζn : n < ωi is a strictly increasing sequence from RM (N). Then by deﬁnition, for each n > 0 there is γn ∈ M such that ζn = min((N ∩ κ) \ γn). Let ζ := sup{ζn : n < ω}. Then ζ = sup{γn : n < ω}. Therefore ζ ∈ cl(M ∩ κ) ∩ cl(N ∩ κ).

Hence ζ < βM,N by Lemma 1.15. But

βM,N ≤ ζ0 < ζ, which is a contradiction. (3) Since ζ > min(RM (N) ∪ RN (M)) and RM (N) and RN (M) are ﬁnite, let σ0 be the largest member of RM (N)∪RN (M) less than ζ. We claim that σ0 ∈ RN (M). If not, then σ0 ∈ RM (N), and in particular, σ0 ∈ (N ∩ζ)\βM,N . Since ζ ∈ RM (N), by the deﬁnition of RM (N) we have that M ∩(σ0, ζ) 6= ∅. But then min((M ∩κ)\σ0) is in RN (M) and is between σ0 and ζ, which contradicts the maximality of σ0. We claim that ζ = min((N ∩ κ) \ σ0). Otherwise min((N ∩ κ) \ σ0) is in RM (N) and is between σ0 and ζ, which contradicts the maximality of σ0. It follows that sup(N ∩ ζ) ≤ σ0. Finally, we show that σ0 = min((M ∩ κ) \ sup(N ∩ ζ)). Therefore σ = σ0, and we are done. Suppose for a contradiction that σ < σ0. As sup(N ∩ζ) ≤ σ, we have that N ∩ (σ, σ0) = ∅. Observe that βM,N ≤ σ. For if σ < βM,N , then σ ∈ (M∩βM,N )\N, which implies that N < M. And since sup(N ∩ ζ) ≤ σ, it follows that ζ = min((N ∩ κ) \ βM,N ). So ζ = min(RM (N) ∪ RN (M)), which is a contradiction. Hence βM,N ≤ σ < σ0. Since σ0 ∈ RN (M), there is γ ∈ N such that σ0 = min((M ∩ κ) \ γ). But then σ < γ < σ0, which contradicts that N ∩ (σ, σ0) = ∅.

The rest of the section follows roughly the same sequence of topics covered in the previous section. Lemma 2.3 describes the remainder points which appear when adding M ∩N to an adequate set, where M < N, as in Lemma 1.24 and Proposition 1.25. Then Lemmas 2.4–2.6 analyze remainder points which appear in the process of amalgamating over countable models, as in Proposition 1.29.

Lemma 2.3. Let K, M, and N be in X0. Suppose that M < N and {K,M,N} is adequate. Then:

(1) RK (M ∩ N) ⊆ RK (M); (2) RM∩N (K) ⊆ RM (K) ∪ RN (K).

Proof. Note that by Lemma 1.24, {K,M ∩ N} is adequate, βK,M∩N ≤ βK,M , and βK,M∩N ≤ βM,N . (1) Let ζ ∈ RK (M ∩ N), and we will show that ζ ∈ RK (M). Then either (a) K ∼ M ∩N and ζ = min((M ∩N ∩κ)\βK,M∩N ), or (b) there is γ ∈ (K∩κ)\βK,M∩N such that ζ = min((M ∩ N ∩ κ) \ γ).

Case a: K ∼ M ∩N and ζ = min((M ∩N ∩κ)\βK,M∩N ). Then by Lemma 1.24, K ∼ M. We claim that βK,M ≤ ζ. Suppose for a contradiction that ζ < βK,M . Then since K ∼ M and ζ ∈ M ∩ βK,M , it follows that ζ ∈ K. But this contradicts that ζ ∈ RK (M ∩ N). MITCHELL’S THEOREM REVISITED 15

Since βK,M∩N ≤ βK,M ≤ ζ, it follows that ζ = min((M ∩ N ∩ κ) \ βK,M ). As M < N, M ∩ N ∩ κ = M ∩ βM,N , which is an initial segment of M ∩ κ. So ζ = min((M ∩ κ) \ βK,M ), and hence ζ ∈ RK (M).

Case b: There is γ ∈ (K ∩ κ) \ βK,M∩N such that ζ = min((M ∩ N ∩ κ) \ γ). Since M ∩ N ∩ κ = M ∩ βM,N is an initial segment of M ∩ κ, it follows that ζ = min((M ∩ κ) \ γ). If βK,M ≤ γ, then since γ ∈ K, ζ ∈ RK (M). So assume that γ < βK,M . Now ζ ∈ M ∩ N ∩ κ implies that ζ < βM,N . So γ < βM,N . Since γ ∈ (K ∩ κ) \ βK,M∩N , γ∈ / M ∩ N. But M ∩ N ∩ κ = M ∩ βM,N , so γ∈ / M ∩ κ. Since γ < βK,M and γ ∈ K \ M, we have that M < K. So M ∩ βK,M ⊆ K. As ζ ∈ RK (M ∩ N), ζ∈ / K. Since M ∩ βK,M ⊆ K and ζ ∈ M \ K, it follows that βK,M ≤ ζ. In conclusion, γ < βK,M ≤ ζ. Hence ζ = min((M ∩ κ) \ βK,M ). Since M < K, this implies that ζ ∈ RK (M).

(2) Let ζ ∈ RM∩N (K). Then either (a) K ∼ M ∩ N and ζ = min((K ∩ κ) \ βK,M∩N ), or (b) there is γ ∈ (M ∩ N) \ βK,M∩N such that ζ = min((K ∩ κ) \ γ). We will show that either ζ ∈ RM (K) or ζ ∈ RN (K).

Case a: K ∼ M ∩ N and ζ = min((K ∩ κ) \ βK,M∩N ). Then K ∼ M by Lemma 1.24. Assume ﬁrst that βK,M ≤ ζ. Then βK,M∩N ≤ βK,M ≤ ζ. So ζ = min((K ∩ κ) \ βK,M ), which implies that ζ ∈ RM (K). Now assume that ζ < βK,M . Since K ∼ M and ζ ∈ K ∩ βK,M , ζ ∈ M. As ζ ∈ RM∩N (K), ζ∈ / M ∩ N, so ζ∈ / N. Since K ∼ M < N, K < N. As ζ ∈ K \ N and K < N, βK,N ≤ ζ. Since M ∩ N ⊆ N, βK,M∩N ≤ βK,N . Hence

βK,M∩N ≤ βK,N ≤ ζ.

So ζ = min((K ∩ κ) \ βK,N ), and therefore ζ ∈ RN (K).

Case b: ζ = min((K ∩ κ) \ γ), for some γ ∈ (M ∩ N) \ βK,M∩N . If βK,M ≤ γ, then γ ∈ (M ∩ κ) \ βK,M , and hence ζ ∈ RM (K). Suppose that γ < βK,M ≤ ζ. Then ζ = min((K ∩ κ) \ βK,M ). Since γ ∈ (M ∩ βK,M ) \ K, K < M. Therefore ζ ∈ RM (K). The remaining case is that γ < ζ < βK,M . Since βK,M∩N ≤ γ and γ ∈ M ∩ N, γ∈ / K. So γ ∈ (M ∩ βK,M ) \ K. It follows that K < M. But ζ ∈ K ∩ βK,M , so ζ ∈ M. As ζ ∈ RM∩N (K) and ζ ∈ M, ζ∈ / N. But K < M < N, so K < N. As ζ ∈ K \ N, βK,N ≤ ζ. If βK,N ≤ γ, then γ ∈ (N ∩ κ) \ βK,N , and therefore ζ ∈ RN (K). Suppose that γ < βK,N ≤ ζ. Then ζ = min((K ∩ κ) \ βK,N ). Since K < N, ζ ∈ RN (K). Lemmas 2.4 and 2.5 describe the same situation we considered in Lemmas 1.26 and 1.27.

Lemma 2.4. Let N ≤ M and L ∈ N, where L, M, and N are in X0. Then:

(1) for all ζ ∈ RL(M), βM,N ≤ ζ and ζ ∈ RN (M); (2) for all ζ ∈ RM (L), there is ξ ∈ RM (N) such that ζ = min((L ∩ κ) \ ξ). Proof. Note that by Lemma 1.26, L < M. (1) Let ζ ∈ RL(M). Since L < M, there is γ ∈ (L ∩ κ) \ βL,M such that ζ = min((M ∩ κ) \ γ). Since γ ∈ L and L ∈ N, γ ∈ N. So γ ∈ N \ M. Since N ≤ M, βM,N ≤ γ. Hence βM,N ≤ ζ. As ζ = min((M ∩ κ) \ γ), ζ ∈ RN (M). 16 THOMAS GILTON AND JOHN KRUEGER

(2) Let ζ ∈ RM (L). Since L < M, there is γ ∈ (M ∩ κ) \ βL,M such that ζ = min((L ∩ κ) \ γ). Now ζ ∈ L \ M and L ∈ N. So ζ ∈ N \ M. Since N ≤ M, this implies that βM,N ≤ ζ. If γ < βM,N , then let ξ := min((N ∩ κ) \ βM,N ). Since N ≤ M, ξ ∈ RM (N). As L ⊆ N, clearly ζ = min((L∩κ)\ξ). If βM,N ≤ γ, then let ξ := min((N∩κ)\γ), which exists since ζ ∈ N. Then ξ ∈ RM (N), and since L ⊆ N, ζ = min((L ∩ κ) \ ξ).

Lemma 2.5. Let M < N and L ∈ N, where L, M, and N are in X0. Then:

(1) for all ζ ∈ RL(M), either ζ < βM,N and ζ ∈ RL(M ∩ N), or βM,N ≤ ζ and ζ ∈ RN (M); (2) for all ζ ∈ RM (L), either ζ ∈ RM∩N (L) or there is ξ ∈ RM (N) such that ζ = min((L ∩ κ) \ ξ).

Proof. Note that by Lemma 1.27, βL,M = βL,M∩N . And since M < N, M ∩βM,N = M ∩ N ∩ κ. (1) Let ζ ∈ RL(M). Then either (a) L ∼ M and ζ = min((M ∩ κ) \ βL,M ), or (b) there is γ ∈ (L ∩ κ) \ βL,M such that ζ = min((M ∩ κ) \ γ). Assume ﬁrst that ζ < βM,N . In case (a), L ∼ M ∩ N by Lemma 1.27. Since ζ < βM,N , ζ = min((M ∩ N ∩ κ) \ βL,M∩N ). In case (b), γ ∈ (L ∩ κ) \ βL,M∩N and ζ = min((M ∩ N ∩ κ) \ γ). In either case, ζ ∈ RL(M ∩ N). Now assume that βM,N ≤ ζ. In case (a), since

βL,M ≤ βM,N ≤ ζ,

ζ = min((M ∩κ)\βM,N ). Since M < N, this implies that ζ ∈ RN (M). In case (b), if γ < βM,N , then again ζ = min((M ∩ κ) \ βM,N ), and so ζ ∈ RN (M). Otherwise γ ∈ (N ∩ κ) \ βM,N and ζ = min((M ∩ κ) \ γ), so ζ ∈ RN (M).

(2) Let ζ ∈ RM (L). Then either (a) L ∼ M and ζ = min((L ∩ κ) \ βL,M ), or (b) there is γ ∈ (M ∩ κ) \ βL,M such that ζ = min((L ∩ κ) \ γ). In case (a), L ∼ M ∩ N by Lemma 1.27 and ζ = min((L ∩ κ) \ βL,M∩N ). Hence ζ ∈ RM∩N (L). Assume (b). First consider the case that γ < βM,N . Then

γ ∈ M ∩ βM,N ⊆ M ∩ N. So

γ ∈ (M ∩ N ∩ κ) \ βL,M∩N and ζ = min((L ∩ κ) \ γ). Hence ζ ∈ RM∩N (L). Now consider the case that βM,N ≤ γ. Then γ ∈ (M ∩ κ) \ βM,N . Let ξ := min((N ∩ κ) \ γ), which exists since ζ ∈ N. Then ξ ∈ RM (N) and ζ = min((L ∩ κ) \ ξ). When amalgamating over a countable model N, the presence of M ∩ N prevents certain incompatibilities between M and the object we build in N. But oftentimes M ∩ N does not have enough information about M. In that case, we will use a model M 0 in N which is more representative of M than M ∩ N.

0 Lemma 2.6. Let L, M, M , and N be in X0. Assume that M < N and L ∈ N. 0 0 0 Also suppose that M ∈ N, {L, M ∩N,M } is adequate, and M ∩βM,N = M ∩βM,N . Then:

(1) either βL,M = βL,M 0 or βM,N < βL,M 0 ; (2) if βL,M = βL,M 0 and ζ ∈ RM∩N (L), then ζ ∈ RM 0 (L). MITCHELL’S THEOREM REVISITED 17

Proof. Note that {L, M} is adequate by Lemma 1.28. We claim that βL,M ≤ βL,M 0 . Otherwise βL,M 0 < βL,M . Since {L, M} is adequate, we can ﬁx ξ ∈ (L ∩ M) ∩ [βL,M 0 , βL,M ) by Lemma 1.19(5). Since L ∈ N, ξ ∈ N. So 0 ξ ∈ M ∩ N ∩ κ = M ∩ βM,N ⊆ M . 0 Hence ξ ∈ (L ∩ M ∩ κ) \ βL,M 0 , which is impossible. (1) If βL,M = βL,M 0 , then we are done. So assume that βL,M < βL,M 0 . We claim that βM,N < βL,M 0 . Otherwise

βL,M < βL,M 0 ≤ βM,N . 0 0 Since {L, M } is adequate, we can ﬁx ξ ∈ (L ∩ M ) ∩ [βL,M , βL,M 0 ) by Lemma 1.19(5). Then 0 ξ ∈ M ∩ βM,N ⊆ M.

So ξ ∈ (L ∩ M ∩ κ) \ βL,M , which is a contradiction. (2) Assume that βL,M = βL,M 0 and ζ ∈ RM∩N (L). By Lemma 1.27,

βL,M 0 = βL,M = βL,M∩N .

First, assume that L ∼ M ∩ N and ζ = min((L ∩ κ) \ βL,M∩N ). Then ζ = min((L ∩ κ) \ βL,M 0 ). Also 0 0 L ∩ ω1 = (M ∩ N) ∩ ω1 = M ∩ βM,N ∩ ω1 = M ∩ βM,N ∩ ω1 = M ∩ ω1. 0 0 0 Since {L, M } is adequate and L ∩ ω1 = M ∩ ω1, L ∼ M by Lemma 1.21. Since 0 L ∼ M and ζ = min((L ∩ κ) \ βL,M 0 ), ζ ∈ RM 0 (L). Secondly, suppose that γ ∈ (M ∩ N ∩ κ) \ βL,M∩N and ζ = min((L ∩ κ) \ γ). Then

γ ∈ (M ∩ N ∩ κ) \ βL,M 0 . Since 0 M ∩ N ∩ κ = M ∩ βM,N ⊆ M , 0 γ ∈ (M ∩ κ) \ βL,M 0 . So ζ ∈ RM 0 (L). The statement of the next technical lemma is not very intuitive. But its discovery led to substantial simpliﬁcations of some of the arguments from [8].

Lemma 2.7. Let K, M, and N be in X0 such that {K,M,N} is adequate. Suppose that

ζ ∈ RM (N), ζ∈ / K, θ = min((K ∩ κ) \ ζ), and θ < βK,N .

Then θ ∈ RM (K).

Proof. Since ζ < θ < βK,N and ζ ∈ N \ K, it follows that K < N. In particular, K ∩ (θ + 1) ⊆ N.

Case 1: N ≤ M. Then K < N ≤ M, so K < M. We claim that βK,M ≤ βM,N . Otherwise βM,N < βK,M , which implies that

(K ∩ M) ∩ [βM,N , βK,M ) 6= ∅ by Lemma 1.19(5). Let γ = min((K ∩κ)\βM,N ). Then since the intersection above is nonempty, γ < βK,M , and hence γ ∈ K ∩ M. But βM,N ≤ ζ < θ and θ ∈ K implies that γ ≤ θ. Since K ∩ (θ + 1) ⊆ N, γ ∈ N. So γ ∈ (M ∩ N) \ βM,N , which is impossible. This proves that βK,M ≤ βM,N . 18 THOMAS GILTON AND JOHN KRUEGER

Suppose that ζ = min((N ∩ κ) \ γ) for some γ ∈ (M ∩ κ) \ βM,N . Since βK,M ≤ βM,N , it follows that γ ∈ (M ∩κ)\βK,M . As K ∩(θ +1) ⊆ N, θ = min((K ∩κ)\γ). Hence θ ∈ RM (K). Suppose that M ∼ N and ζ = min((N ∩ κ) \ βM,N ). Since K ∩ (θ + 1) ⊆ N, it follows that θ = min((K ∩κ)\βM,N ). We claim that θ = min((K ∩κ)\βK,M ), which implies that θ ∈ RM (K) as desired. If not, then there is π ∈ K ∩ [βK,M , βM,N ). But βM,N ≤ ζ < θ < βK,N , so π ∈ K ∩ βK,N ⊆ N. Hence π ∈ N ∩ βM,N ⊆ M. So π ∈ M. Therefore π ∈ (K ∩ M) \ βK,M , which is impossible.

Case 2: M < N. Since ζ ∈ RM (N), there is γ ∈ (M ∩ κ) \ βM,N such that ζ = min((N∩κ)\γ). If βK,M ≤ γ, then γ ∈ (M∩κ)\βK,M , and since K∩(θ+1) ⊆ N, θ = min((K ∩ κ) \ γ). Hence θ ∈ RM (K). Otherwise γ < βK,M . Since γ∈ / N, γ < θ, and K ∩ (θ + 1) ⊆ N, it follows that γ∈ / K. So γ ∈ (M ∩βK,M )\K, which implies that K < M. Since K ∩(θ +1) ⊆ N, it follows that θ = min((K ∩ κ) \ γ). As θ ∈ (N ∩ κ) \ βM,N , θ∈ / M. As K < M and θ ∈ K ∩ κ, βK,M ≤ θ. So γ < βK,M ≤ θ. Hence θ = min((K ∩ κ) \ βK,M ), which implies that θ ∈ RM (K).

The next three lemmas are analogues of Lemmas 2.3, 2.5, and 2.6, where the countable model N in X0 is replaced by an uncountable model P in Y0.

Lemma 2.8. Let K and M be in X0 and P ∈ Y0. Assume that {K,M} is adequate and sup(M ∩ P ∩ κ) < P ∩ κ. Then:

(1) RK (M ∩ P ) ⊆ RK (M); (2) if ζ ∈ RM∩P (K), then either ζ ∈ RM (K) or ζ = min((K ∩ κ) \ (P ∩ κ)).

Proof. Note that by Lemma 1.32, βK,M∩P ≤ βK,M , βK,M∩P < P ∩ κ, and {K,M ∩ P } is adequate.

(1) Let ζ ∈ RK (M ∩ P ). Then either (a) K ∼ M ∩ P and ζ = min((M ∩ P ∩ κ) \ βK,M∩P ), or (b) there is γ ∈ (K ∩ κ) \ βK,M∩P such that ζ = min((M ∩ P ∩ κ) \ γ).

Case a: K ∼ M ∩ P and ζ = min((M ∩ P ∩ κ) \ βK,M∩P ). Then K ∼ M by Lemma 1.32. By Lemma 1.36, either βK,M = βK,M∩P , or P ∩ κ < βK,M . We claim that βK,M = βK,M∩P . Suppose for a contradiction that P ∩κ < βK,M . Since ζ ∈ M ∩ P ∩ κ ⊆ P ∩ κ, ζ < βK,M . But since K ∼ M and ζ ∈ M ∩ βK,M , ζ ∈ K. So ζ ∈ K ∩ (M ∩ P ) ∩ κ, which contradicts that ζ ∈ RK (M ∩ P ). So βK,M = βK,M∩P . Since M ∩ P ∩ κ is an initial segment of M ∩ κ, ζ = min((M ∩ κ) \ βK,M ). Hence ζ ∈ RK (M).

Case b: ζ = min((M ∩ P ∩ κ) \ γ), for some γ ∈ (K ∩ κ) \ βK,M∩P . Since M ∩ P ∩ κ is an initial segment of M ∩ κ, ζ = min((M ∩ κ) \ γ). By Lemma 1.36, either βK,M = βK,M∩P or P ∩ κ < βK,M . In the ﬁrst case, γ ∈ (K ∩ κ) \ βK,M , so ζ ∈ RK (M). We prove that the other case is impossible. Suppose for a contradiction that P ∩ κ < βK,M . Since γ < ζ < P ∩ κ, γ ∈ P . But γ ∈ (K ∩ κ) \ βK,M∩P implies that γ∈ / M ∩ P . So γ∈ / M. As γ < P ∩ κ < βK,M , we have that γ ∈ (K ∩ βK,M ) \ M. Hence M < K. Since ζ ∈ M ∩ P ∩ κ, ζ ∈ M ∩ βK,M . As M < K, ζ ∈ K. But this is impossible since ζ ∈ RK (M ∩ P ). MITCHELL’S THEOREM REVISITED 19

(2) Let ζ ∈ RM∩P (K). We will prove that either ζ ∈ RM (K), or ζ = min((K ∩ κ) \ (P ∩ κ)). Either (a) K ∼ M ∩ P and ζ = min((K ∩ κ) \ βK,M∩P ), or (b) there is γ ∈ (M ∩ P ∩ κ) \ βK,M∩P such that ζ = min((K ∩ κ) \ γ).

Case a: K ∼ M ∩ P and ζ = min((K ∩ κ) \ βK,M∩P ). Then K ∼ M by Lemma 1.32. Also by Lemma 1.36, either βK,M = βK,M∩P or P ∩ κ < βK,M . First, assume that βK,M = βK,M∩P . Then ζ = min((K ∩ κ) \ βK,M ), so ζ ∈ RM (K). Secondly, assume that P ∩ κ < βK,M . Suppose that βK,M ≤ ζ. Since βK,M∩P ≤ βK,M , it follows that ζ = min((K ∩ κ) \ βK,M ). Therefore ζ ∈ RM (K). Otherwise ζ < βK,M . But then K ∼ M and ζ ∈ K ∩ βK,M imply that ζ ∈ M. Since ζ ∈ RM∩P (K) and ζ ∈ M, ζ∈ / P ∩ κ. Therefore βK,M∩P < P ∩ κ ≤ ζ. So ζ = min((K ∩ κ) \ (P ∩ κ)).

Case b: ζ = min((K ∩ κ) \ γ) for some γ ∈ (M ∩ P ∩ κ) \ βK,M∩P . If P ∩ κ ≤ ζ, then γ < P ∩ κ ≤ ζ implies that ζ = min((K ∩ κ) \ (P ∩ κ)). Suppose that ζ < P ∩ κ. If βK,M ≤ γ, then γ ∈ (M ∩ κ) \ βK,M , and therefore ζ ∈ RM (K). So assume that γ < βK,M . First consider the case that βK,M ≤ ζ. Then ζ = min((K ∩ κ) \ βK,M ). Since γ ∈ (M ∩ βK,M ) \ K, it follows that K < M. So ζ ∈ RM (K). In the ﬁnal case, assume that γ < ζ < βK,M . We will show that this case does not occur. Then βK,M∩P ≤ γ < ζ < βK,M .

Since γ ∈ (M ∩ βK,M ) \ K, it follows that K < M. So as ζ ∈ K ∩ βK,M , ζ ∈ M. But also ζ ∈ P ∩ κ. So ζ ∈ M ∩ P , which contradicts that ζ ∈ RM∩P (K).

Lemma 2.9. Let L and M be in X0 and P in Y0. Assume that L ∈ P , {L, M ∩P } is adequate, and sup(M ∩ P ∩ κ) < P ∩ κ. Then:

(1) if ζ ∈ RL(M), then either ζ ∈ RL(M ∩ P ) or ζ = min((M ∩ κ) \ (P ∩ κ)); (2) RM (L) ⊆ RM∩P (L).

Proof. Note that by Lemma 1.34, βL,M = βL,M∩P , βL,M < P ∩ κ, and {L, M} is adequate.

(1) Let ζ ∈ RL(M). Then either (a) L ∼ M and ζ = min((M ∩ κ) \ βL,M ), or (b) there is γ ∈ (L ∩ κ) \ βL,M such that ζ = min((M ∩ κ) \ γ).

Case a: L ∼ M and ζ = min((M ∩κ)\βL,M ). Then L ∼ M ∩P by Lemma 1.34. If P ∩ κ ≤ ζ, then since βL,M < P ∩ κ, it follows that ζ = min((M ∩ κ) \ (P ∩ κ)). Suppose that ζ < P ∩ κ. Then

ζ = min((M ∩ P ∩ κ) \ βL,M ) = min((M ∩ P ∩ κ) \ βL,M∩P ).

So ζ ∈ RL(M ∩ P ).

Case b: There is γ ∈ (L ∩ κ) \ βL,M such that ζ = min((M ∩ κ) \ γ). Then γ ∈ (L∩κ)\βL,M∩P . If ζ < P ∩κ, then ζ = min((M ∩P ∩κ)\γ), so ζ ∈ RL(M ∩P ). Otherwise P ∩ κ ≤ ζ, and since γ ∈ L, γ < P ∩ κ. So ζ = min((M ∩ κ) \ (P ∩ κ)).

(2) Let ζ ∈ RM (L), and we will show that ζ ∈ RM∩P (L). Either (a) L ∼ M and ζ = min((L ∩ κ) \ βL,M ), or (b) there is γ ∈ (M ∩ κ) \ βL,M such that ζ = min((L ∩ κ) \ γ). 20 THOMAS GILTON AND JOHN KRUEGER

Assume (a). Then L ∼ M ∩ P by Lemma 1.34. Also ζ = min((L ∩ κ) \ βL,M∩P ), so ζ ∈ RM∩P (L). Assume (b). Since ζ ∈ L and L ∈ P , ζ ∈ P . As γ < ζ and ζ ∈ P ∩ κ, γ < P ∩ κ. So γ ∈ M ∩ P . Thus γ ∈ (M ∩ P ∩ κ) \ βL,M∩P and ζ = min((L ∩ κ) \ γ). So ζ ∈ RM∩P (L). 0 0 Lemma 2.10. Let L, M, and M be in X0, and let P and P be in Y0. Assume that {L, M, M 0} is adequate, and L, M 0, and P 0 are in P . Let β := P ∩ κ and β0 := P 0 ∩ κ. Suppose that sup(M ∩ β) < β0 and M ∩ β = M 0 ∩ β0. Then: 0 (1) βL,M < β ; 0 (2) either βL,M = βL,M 0 or β < βL,M 0 ; (3) if βL,M = βL,M 0 and ζ ∈ RM∩P (L), then ζ ∈ RM 0 (L). Proof. (1) Since M ∩ β ⊆ β0 and L ∩ κ ⊆ β, L ∩ M ∩ κ ⊆ β0. As sup(M ∩ β) < β0, L ∩ M ∩ κ is a bounded subset of β0. By the elementarity of P 0, ﬁx γ ∈ Λ such that sup(L ∩ M ∩ κ) < γ < β0. By Lemma 1.19(3), 0 βL,M = min(Λ \ sup(L ∩ M ∩ κ)) ≤ γ < β .

(2) If βL,M = βL,M 0 , then we are done. So suppose not. We claim that βL,M < βL,M 0 . Suppose for a contradiction that βL,M 0 < βL,M . By Lemma 0 1.19(3), βL,M 0 = min(Λ \ sup(L ∩ M ∩ κ)). But 0 βL,M 0 < βL,M < β 0 0 0 0 by (1) and the assumption just made. So βL,M 0 < β . Hence L∩M ∩κ = L∩M ∩β . Since M ∩ β = M 0 ∩ β0, it follows that sup(L ∩ M 0 ∩ κ) = sup(L ∩ M 0 ∩ β0) = sup(L ∩ M ∩ β) = sup(L ∩ M ∩ κ). So

βL,M 0 = min(Λ \ sup(L ∩ M ∩ κ)) = βL,M .

But this contradicts the assumption that βL,M 0 < βL,M . This proves that βL,M < βL,M 0 . By Lemma 1.19(5), we can ﬁx ξ ∈ (L ∩ 0 M ) ∩ [βL,M , βL,M 0 ). Since βL,M ≤ ξ and ξ ∈ L, it follows that ξ∈ / M. But 0 0 0 0 M ∩ β = M ∩ β . Since ξ ∈ (M ∩ κ) \ M, β ≤ ξ. As ξ < βL,M 0 , it follows that 0 β < βL,M 0 . (3) Suppose that βL,M = βL,M 0 and ζ ∈ RM∩P (L). We will prove that ζ ∈ RM 0 (L). Since βL,M = βL,M∩P by Lemma 1.34, βL,M 0 = βL,M∩P . First assume that L ∼ M ∩ P and ζ = min((L ∩ κ) \ βL,M∩P ). Hence ζ = min((L ∩ κ) \ βL,M 0 ). As L ∼ M ∩ P , 0 L ∩ ω1 = M ∩ P ∩ ω1 = M ∩ ω1. 0 So L ∼ M by Lemma 1.21. Hence ζ ∈ RM 0 (L). Now assume that ζ = min((L ∩ κ) \ γ), where γ ∈ (M ∩ P ∩ κ) \ βL,M∩P . Since M ∩ P ∩ κ = M ∩ β ⊆ M 0, γ ∈ M 0. And

βL,M∩P = βL,M = βL,M 0 ≤ γ. 0 So γ ∈ (M ∩ κ) \ βL,M 0 . Therefore ζ ∈ RM 0 (L).

The ﬁnal lemma concerning remainder points will be used when amalgamating over transitive models. MITCHELL’S THEOREM REVISITED 21

0 0 0 Lemma 2.11. Let M, M , N, and N be in X0. Assume that M ∩ κ = M ∩ κ and 0 0 N ∩ κ = N ∩ κ. Then RM (N) = RM 0 (N ). 0 Proof. We will show that RM (N) ⊆ RM 0 (N ). The reverse inclusion follows by symmetry. So let ζ ∈ RM (N). First, assume that M ∼ N and ζ = min((N ∩ κ) \ βM,N ). Then by Lemma 1.37, 0 0 0 0 βM,N = βM 0,N 0 and M ∼ N . Since N ∩ κ = N ∩ κ, clearly ζ = min((N ∩ κ) \ 0 βM 0,N 0 ). So ζ ∈ RM 0 (N ). Secondly, assume that ζ = min((N ∩ κ) \ γ), for some γ ∈ (M ∩ κ) \ βM,N . By 0 0 Lemma 1.37, βM,N = βM 0,N 0 . Since M ∩ κ = M ∩ κ, γ ∈ (M ∩ κ) \ βM 0,N 0 . As 0 0 0 N ∩ κ = N ∩ κ, ζ = min((N ∩ κ) \ γ). So ζ ∈ RM 0 (N ).

§3. Strong genericity and cardinal preservation

In this section we will discuss the idea of a strongly generic condition, which is due to Mitchell [12]. Then we will use the existence of strongly generic conditions to prove cardinal preservation results. All of the results in this section are either due to Mitchell, or are based on standard proper forcing arguments.

Deﬁnition 3.1. Let Q be a forcing poset, q ∈ Q, and N a set. We say that q is a strongly N-generic condition if for any set D which is a dense subset of N ∩ Q, D is predense in Q below q. Note that if q is strongly N-generic and r ≤ q, then r is strongly N-generic.

Notation 3.2. For a forcing poset Q, let λQ denote the least cardinal such that Q ⊆ H(λQ).

Note that q is strongly N-generic iﬀ q is strongly (N ∩ H(λQ))-generic. The following proposition gives a more intuitive description of strong genericity.

Lemma 3.3. Let Q be a forcing poset, q ∈ Q, and N ≺ (H(χ), ∈, Q), where λQ ≤ χ is a cardinal. Then q is a strongly N-generic condition iff q forces that N ∩ G˙ is a V -generic filter on N ∩ Q. Proof. Suppose that q is a strongly N-generic condition, and let G be a V -generic filter on Q containing q. We will show that N ∩ G is a V -generic filter on N ∩ Q. First, we show that N ∩ G is a filter on N ∩ Q. If p ∈ N ∩ G and t ∈ N ∩ Q with p ≤ t, then t ∈ G since G is a filter, and hence t ∈ N ∩ G. Suppose that s and t are in N ∩ G, and we will find p ∈ N ∩ G such that p ≤ s, t. The set D of p in N ∩ Q which are either incompatible with one of s and t, or below both s and t, is a dense subset of N ∩ Q by the elementarity of N. Since q is strongly N-generic, D is predense below q. As q ∈ G and G is a V -generic filter, we can fix p ∈ G ∩ D. Since s, t, and p are in G, p is compatible with s and t, and therefore p ≤ s, t by the definition of D. As D ⊆ N, p ∈ N ∩ G. Secondly, we prove that N ∩G is V -generic on N ∩Q. So let D be a dense subset of N ∩ Q. Since q is a strongly N-generic condition, D is predense below q. As q ∈ G, it follows that D ∩ G 6= ∅. But D ⊆ N, so D ∩ N ∩ G 6= ∅. Conversely, suppose that q forces that N ∩ G˙ is a V -generic filter on N ∩ Q, and we will show that q is strongly N-generic. Let D be a dense subset of N ∩ Q. If D is not predense below q, then we can fix r ≤ q which is incompatible with every condition in D. Let G be a V -generic filter on Q containing r. Since r ≤ q, q ∈ G. 22 THOMAS GILTON AND JOHN KRUEGER

Hence by assumption, N ∩ G is a V -generic filter on N ∩ Q. Since D is dense in N ∩ Q, we can fix s ∈ G ∩ D. Then r is incompatible with s by the choice of r, and yet r and s are compatible since they are both in the filter G. The following combinatorial characterization of strong genericity is very useful in practice.

Lemma 3.4. Let Q be a forcing poset, q ∈ Q, and N a set. Then the following are equivalent: (1) q is strongly N-generic; (2) for all r ≤ q, there exists v ∈ N ∩ Q such that for all w ≤ v in N ∩ Q, r and w are compatible. Proof. For the forward direction, suppose that there is r ≤ q for which there does not exist a condition v ∈ N ∩Q all of whose extensions in N ∩Q are compatible with r. Let D be the set of w ∈ N ∩ Q which are incompatible with r. The assumption on r implies that D is dense in N ∩ Q. But D is not predense below q since every condition in D is incompatible with r. So q is not strongly N-generic. Conversely, assume that there is a function r 7→ vr as described in (2). Let D be dense in N ∩ Q, and let r ≤ q. Since D is dense in N ∩ Q, we can ﬁx w ≤ vr in D. Then r and w are compatible by the choice of vr. So D is predense below q.

The next idea was introduced by Cox-Krueger [2].

Deﬁnition 3.5. Let Q be a forcing poset, q ∈ Q, and N a set. We say that q is a universal strongly N-generic condition if q is a strongly N-generic condition and for all p ∈ N ∩ Q, p and q are compatible. The strongly generic conditions used in this paper are universal. This fact allows us to factor forcing posets over elementary substructures in such a way that the quotient forcing has nice properties. See Section 6 for more details on this topic.

Deﬁnition 3.6. Let Q be a forcing poset and µ ≤ λQ a regular uncountable cardinal. We say that Q is µ-strongly proper on a stationary set if there are stationarily many N in Pµ(H(λQ)) such that for all p ∈ N ∩ Q, there is q ≤ p such that q is strongly N-generic.

When we say that Q is strongly proper on a stationary set, we will mean that it is ω1-strongly proper on a stationary set. By standard arguments, Q is µ-strongly proper on a stationary set iﬀ for any cardinal λQ ≤ χ, there are stationarily many N in Pµ(H(χ)) such that for all p ∈ N ∩ Q, there is q ≤ p such that q is strongly N-generic.

Lemma 3.7. Let Q be a forcing poset and µ ≤ λQ a regular uncountable cardinal. If there are stationarily many N in Pµ(H(λQ)) such that there exists a universal strongly N-generic condition, then Q is µ-strongly proper on a stationary set.

Proof. Let N ∈ Pµ(H(λQ)) be such that there exists a universal strongly N-generic condition qN . Let p ∈ N ∩ Q, and we will ﬁnd r ≤ p which is strongly N-generic. Since qN is universal, p and qN are compatible. So ﬁx r ≤ p, qN . Then r ≤ p and r is strongly N-generic. MITCHELL’S THEOREM REVISITED 23

Deﬁnition 3.8. Let µ be a regular uncountable cardinal. A forcing poset Q is said to satisfy the µ-covering property if Q forces that for any set a ⊆ On in the generic extension, if a has size less than µ in the generic extension, then there is b in the ground model with size less than µ in the ground model such that a ⊆ b.

Note that if Q has the µ-covering property, then Q forces that µ is regular.

Proposition 3.9. Let Q be a forcing poset, and let µ ≤ λQ be a regular uncountable cardinal. Suppose that Q is µ-strongly proper on a stationary set. Then Q satisﬁes the µ-covering property.1 Proof. Let p be a condition, and suppose that p forces thata ˙ is a set of ordinals of size less than µ. We will ﬁnd q ≤ p and a set x of size less than µ such that q forces thata ˙ ⊆ x. Extending p if necessary, we can assume that p forces thata ˙ has size µ0, for some cardinal µ0 < µ. Fix a sequence hα˙ i : i < µ0i of Q-names such that p forces thata ˙ = {α˙ i : i < µ0}.

Fix a regular cardinal λQ ≤ χ such that Q,a ˙, and hα˙ i : i < µ0i are members of H(χ). Fix N ∈ Pµ(H(χ)) such that N ≺ (H(χ), ∈, Q, p, a,˙ hα˙ i : i < µ0i), µ0 ⊆ N, and for all p0 ∈ N ∩ Q, there is q ≤ p0 which is a strongly N-generic condition. In particular, since p ∈ N ∩ Q, we can ﬁx q ≤ p such that q is strongly N-generic. We claim that q forces that for all i < µ0,α ˙ i ∈ N. Let i < µ0. Let D be the set of s ∈ N ∩ Q such that s decides the value ofα ˙ i. By the elementarity of N, it is easy to see that D is dense in N ∩ Q. Since q is strongly N-generic, D is predense below q. Therefore q forces thatα ˙ i is decided by a condition in N. By elementarity, the value of the nameα ˙ i decided by a condition in N lies in N. Hence q forces that α˙ i ∈ N. It follows that q forces thata ˙ ⊆ N ∩ On. Since N has size less than µ, we are done.

Corollary 3.10. Let Q be a forcing poset, and let µ ≤ λQ be a regular uncountable cardinal. Suppose that there are stationarily many N in Pµ(H(λQ)) for which there exists a universal strongly N-generic condition. Then Q satisﬁes the µ-covering property. In particular, Q forces that µ is a regular cardinal. Proof. Immediate from Lemma 3.7 and Proposition 3.9.

Proposition 3.11. Let Q be a forcing poset, and let µ ≤ λQ be a regular uncountable cardinal. Suppose that there are stationarily many N ∈ Pµ(H(λQ)) such that every condition in Q is a strongly N-generic condition. Then Q is µ-c.c. Note that if Q has a maximum condition, then every condition in Q being strongly N-generic is equivalent to the maximum condition being strongly N- generic.

Proof. Let A be a maximal antichain in Q, and we will show that |A| < µ. Let N ∈ Pµ(H(λQ)) be such that N ≺ (H(λQ), ∈, Q,A) and every condition in Q is strongly N-generic. Note that by the elementarity of N and since A is a maximal antichain, N ∩ A is predense in N ∩ Q. Namely, if u ∈ N ∩ Q, then u is compatible with some member of A. By elementarity, u is compatible with some member of N ∩ A. Let D be the

1The proof of this proposition is basically the same as a standard proof that proper forcing posets preserve ω1. 24 THOMAS GILTON AND JOHN KRUEGER set of t ∈ N ∩ Q such that for some s ∈ N ∩ A, t ≤ s. Then easily D is dense in N ∩ Q. Since every condition in Q is strongly N-generic, D is predense in Q. We claim that A ⊆ N, which implies that |A| ≤ |N| < µ. So let p ∈ A be given, and we will show that p ∈ N. Since D is predense in Q, fix t ∈ D which is compatible with p. By the definition of D, we can fix s ∈ N ∩ A such that t ≤ s. Then p and s are compatible. But p and s are both in A and A is an antichain, so p = s. Since s ∈ N, p ∈ N. In general, forcings which include adequate sets as side conditions will collapse κ to become ω2. In other words, all the cardinals µ with ω1 < µ < κ will be collapsed to have size ω1. The next result describes some general properties of a forcing poset which imply that such collapsing takes place.

Proposition 3.12. Suppose that Q is a forcing poset which preserves ω1 and sat- isﬁes: (1) there exists an integer k < ω such that the conditions of Q are of the form (x1, . . . , xk,A), where x1, . . . , xk are ﬁnite subsets of H(λ), and A is an adequate set; (2) if (y1, . . . , yk,B) ≤ (x1, . . . , xk,A), then A ⊆ B; (3) there are stationarily many N ∈ X0 such that whenever (x1, . . . , xk,A) ∈ N ∩ Q, then (x1, . . . , xk,A ∪ {N}) is a condition below (x1, . . . , xk,A). Then for any cardinal ω1 < µ < κ, Q collapses µ to have size ω1.

Proof. It suffices to show that Q singularizes all regular cardinals µ with ω1 < µ < κ. For suppose that this is true, but there is a cardinal µ in the interval (ω1, κ) in some generic extension. Assume moreover that µ is the least such cardinal. Then µ = ω2 in the generic extension. By downwards absoluteness, µ is regular in the ground model. This contradicts our assumption that all regular cardinals in the interval (ω1, κ) are singularized. Let G be a V -generic filter on Q. Define

X := {N : ∃(x1, . . . , xk,A) ∈ G, N ∈ A}. Let

Xµ := {N ∈ X : µ ∈ N}.

Then by (2) and the fact that G is a filter, for any M and N in Xµ, there is a condition (x1, . . . , xk,A) ∈ G such that M and N are in A. Since A is adequate, {M,N} is adequate. As µ ∈ M ∩N ∩κ, µ < βM,N . Therefore either M ∩µ = N ∩µ, M ∩ µ ∈ N, or N ∩ µ ∈ M. Moreover, which of these three relations holds is determined by how M ∩ ω1 and N ∩ ω1 are ordered, by Lemma 1.21. It follows that {sup(N ∩ µ): N ∈ Xµ} is a strictly increasing sequence of ordinals with order type at most ω1. We claim that the set {sup(N ∩ µ): N ∈ Xµ} is cofinal in µ. The claim implies that µ has cofinality less than or equal to ω1 in V [G], finishing the proof. Fix a name X˙ µ which is forced to be equal to the set Xµ defined above. Let γ < µ and (x1, . . . , xk,A) be a condition. By (3), there is N ∈ X0 such that (x1, . . . , xk,A), γ, and µ are in N, and (x1, . . . , xk,A ∪ {N}) is a condition below (x1, . . . , xk,A). Since µ ∈ N,(x1, . . . , xk,A ∪ {N}) forces that N ∈ X˙ µ. As γ ∈ N, γ < sup(N ∩ µ). MITCHELL’S THEOREM REVISITED 25

§4. Adding a club

In this section we give an example to illustrate the methods developed so far, by showing how to add a club subset of a stationary subset of ω2 using adequate sets of models. Adding a club with finite conditions was the original application of the side conditions of Friedman [3] and Mitchell [12]. Later Neeman [13] defined a forcing for adding a club using his method of two-type side conditions. The forcing poset we develop in this section is the first example of a forcing which adds a club subset of ω2 using conditions which are just finite sets of models ordered by reverse inclusion. The following general lemma will be used frequently in this section.

Lemma 4.1. Let A be an adequate set. Let K, M, and N be in A, and ζ ∈ RM (N). Suppose that θ = min((K ∩ κ) \ ζ). Then

θ ∈ RM (N) ∪ RM (K) ∪ RN (K).

Proof. If θ = ζ, then θ ∈ RM (N) and we are done. Assume that ζ < θ, which means that ζ∈ / K. If θ < βK,N , then θ ∈ RM (K) by Lemma 2.7. Suppose that βK,N ≤ θ. If βK,N ≤ ζ, then since ζ ∈ N, we have that

θ = min((K ∩ κ) \ ζ) ∈ RN (K).

Otherwise ζ < βK,N ≤ θ. Then θ = min((K ∩κ)\βK,N ). Since ζ ∈ (N ∩βK,N )\K, it follows that K < N. So θ ∈ RN (K). ∗ For the remainder of this section, let κ = λ = ω2. Recall that T is a thin ω1 stationary subset of Pω1 (ω2). We will also assume that 2 = ω2, and hence that ∗ H(ω2) has size ω2. Fix a bijection g : ω2 → H(ω2). Let B denote the structure ∗ ∗ ∗ ∗ (H(ω2), ∈, E,T , π ,C , Λ, g ). ∗ Note that if N is a countable elementary substructure of B and N ∩ ω2 ∈ T , then N ∈ X0. Also note that if M and N are countable elementary substructures of B ∗ and M ∩ ω2 ∈ N, then by the elementarity of M, M = g [M ∩ ω2], and hence by the elementarity of N, M ∈ N. Fix a stationary set S ⊆ ω2 ∩ cof(ω1). We will deﬁne a forcing poset which adds a club subset of S ∪ cof(ω).2

Definition 4.2. A finite set A of elementary substructures of B in X0 is S-adequate if it is adequate, and for all M and N in A, RM (N) ⊆ S. Recall that (B,S) is the structure B augmented with the additional predicate S. Note that the property of being S-adequate is definable in the structure (B,S).

Deﬁnition 4.3. Let P be the forcing poset consisting of S-adequate sets, ordered by reverse inclusion.

We will show that P preserves all cardinals, and adds a club subset of S ∪cof(ω). Note that since H(ω2) has size ω2 and P ⊆ H(ω2), P has size ω2 and thus preserves all cardinals greater than ω2.

2 More generally, it is possible to add a club subset to a fat stationary subset of ω2 using adequate sets as side conditions, but the argument is more complicated than the one which we give here. See [9]. 26 THOMAS GILTON AND JOHN KRUEGER

Proposition 4.4. The forcing poset P is strongly proper on a stationary set. There- fore P satisﬁes the ω1-covering property and preserves ω1.

Proof. Let N be a countable elementary substructure of (B,S) such that N ∩ ω2 ∈ ∗ T . Note that N ∈ X0. Let A0 be in N ∩ P. Deﬁne A1 := A0 ∪ {N}. Observe that A1 is adequate, since for all M ∈ A0, M ∩ βM,N = M ∩ ω2, which is in N. Also A1 is S-adequate, because for all M ∈ A0, RM (N) and RN (M) are empty. Thus A1 is in P and A1 ≤ A0. We claim that A1 is a strongly N-generic condition. By Lemma 3.4, it suﬃces to show that for all A2 ≤ A1, there exists B ∈ N ∩ P such that for all C ≤ B in N ∩ P, A2 ∪ C is S-adequate. Let A2 ≤ A1. We claim that whenever A3 ≤ A2, K and M are in A3, and M < N, then

RM∩N (K) ∪ RK (M ∩ N) ⊆ S.

But this follows immediately from Lemma 2.3 and the fact that A3 is S-adequate. By applying Proposition 1.25 and the last claim ﬁnitely many times, we get that the set A := A2 ∪ {M ∩ N : M ∈ A2, M < N} is S-adequate. Hence A ∈ P and A ≤ A2. Let [ x := {RM (N): M ∈ A}. Since x ⊆ N and x is ﬁnite, x ∈ N. The sets A and N witness that the following statement holds in (B,S):

There are B and N 0 such that B is S-adequate, A ∩ N ⊆ B, N 0 ∈ B, and S 0 x = {RM (N ): M ∈ B}.

The parameters which appear in the above statement, namely A ∩ N and x, are members of N. By the elementarity of N, we can ﬁnd B and N 0 in N which satisfy the same statement. Suppose that C ∈ N ∩ P and C ≤ B. We claim that A ∪ C is S-adequate, which ﬁnishes the proof. Note that if M ∈ A and M < N, then M ∩ N ∈ A. By Lemma 1.19(2), M ∩ βM,N = M ∩ N ∩ ω2. Since M < N, it follows that M ∩ N ∩ ω2 ∈ N. ∗ But M ∩ N = g [M ∩ N ∩ ω2] by the elementarity of M ∩ N, so M ∩ N ∈ N by the elementarity of N. Hence the assumptions of Proposition 1.29 hold. Therefore A ∪ C is adequate. To show that A ∪ C is S-adequate, let M ∈ A and L ∈ C. Let ζ ∈ RL(M), and we will show that ζ ∈ S. By Lemmas 2.4(1) and 2.5(1), we have that

ζ ∈ RN (M) ∪ RL(M ∩ N). Since A and C are S-adequate, M and N are in A, and L and M ∩ N are in C, it follows that ζ ∈ S. Let θ ∈ RM (L). By Lemmas 2.4(2) and 2.5(2), either M < N and θ ∈ RM∩N (L), or there is ξ ∈ RM (N) such that θ = min((L ∩ ω2) \ ξ). In the ﬁrst case, θ ∈ S since C is S-adequate and M ∩ N and L are in C. In the second case, ξ ∈ x, and 0 hence for some K ∈ B, ξ ∈ RK (N ). By Lemma 4.1, 0 θ ∈ RK (N ) ∪ RK (L) ∪ RN 0 (L). 0 Since K, L, and N are in C and C is S-adequate, it follows that θ ∈ S. MITCHELL’S THEOREM REVISITED 27

Lemma 4.5. Suppose that M ∈ X0 and Q is in Y0. Let β := Q ∩ ω2, and assume that cf(β) = ω1 and β ∈ M. Then M ∼ M ∩ Q, RM (M ∩ Q) = ∅, and RM∩Q(M) = {β}.

Proof. By the comments after Notation 1.10, M ∩ Q ∈ X0. By the elementarity of Q, β is a limit point of Λ. Hence the ordinal

β0 := min(Λ \ sup(M ∩ β)) is less than β. By the deﬁnition of β0, clearly

β0 ∈ ΛM ∩ ΛM∩Q.

And since M ∩ Q ∩ ω2 ⊆ β0, β0 is the maximal element of ΛM ∩ ΛM∩Q. Therefore β0 = βM,M∩Q. As M ∩ β0 = M ∩ Q ∩ β0, we have that M ∼ M ∩ Q. Since M ∩ Q ∩ ω2 ⊆ β0 = βM,M∩Q, RM (M ∩ Q) = ∅. As M ∼ M ∩ Q and

β = min((M ∩ ω2) \ β0) = min((M ∩ ω2) \ βM,M∩Q), we have that β ∈ RM∩Q(M). And the fact that M ∩ Q ∩ ω2 ⊆ β0 implies that RM∩Q(M) = {β}. Lemma 4.6. Let Q be an elementary substructure of (B,S) such that Q has size ω1 and Q ∩ ω2 ∈ S. Let β := Q ∩ ω2. Let A0 ∈ Q ∩ P. Suppose that M ∈ X0, and A0 and β are in M. Then β ∈ RM∩Q(M), and

A0 ∪ {M} ∪ {M ∩ Q} is a strongly Q-generic condition.

Proof. Define A1 := A0 ∪ {M}. Then A1 is S-adequate and A1 ≤ A0. Namely, for all K ∈ A0, K ∈ M implies that K ∩ βK,M = K ∩ ω2 ∈ M. So K < M for all K ∈ A0. Also RK (M) and RM (K) are both empty. Define A := A1 ∪{M ∩Q}. By Proposition 1.33, A is adequate. We claim that A is S-adequate. So let K be in A1. If K ∈ A0, then K ∈ M ∩ Q. So RK (M ∩ Q) and RM∩Q(K) are empty. Suppose that K = M. Then by Lemma 4.5, M ∼ M ∩ Q, RM (M ∩ Q) = ∅, and RM∩Q(M) = {β}. Since β ∈ S, we are done. Thus we have established that A is S-adequate. We will show that A is strongly Q-generic. So let A2 ≤ A be given. We claim that for all A3 ≤ A2, for all K ∈ A3, A3 ∪ {K ∩ Q} is S-adequate. By Proposition 1.33, A3 ∪ {K ∩ Q} is adequate. To show that it is S-adequate, let N ∈ A3, and we will show that RN (K ∩ Q) and RK∩Q(N) are subsets of S. By Lemma 2.8(1), RN (K ∩ Q) ⊆ RN (K). Since K and N are in A3 and A3 is S-adequate, RN (K) ⊆ S. Thus RN (K ∩Q) ⊆ S. Now suppose that ζ ∈ RK∩Q(N). Then by Lemma 2.8(2), either ζ ∈ RK (N), or ζ = min((N ∩ ω2) \ β). In the first case, since K and N are in A3, we have that ζ ∈ RK (N) ⊆ S. Assume the second case. Since β ∈ RM∩Q(M) by Lemma 4.5, and M ∩Q, M, and N are in A3, Lemma 4.1 implies that ζ is in RM∩Q(M), RM∩Q(N), or RM (N). Since A3 is S-adequate, ζ ∈ S, which proves the claim. By applying the claim finitely many times, we get that the set ∗ A := A2 ∪ {K ∩ Q : K ∈ A2} is S-adequate. Next we claim that for all K ∈ A2, K ∩ Q is in Q. By Lemma 1.30, K ∩ Q ∩ ω2 ∗ is in Q. Since K and Q are elementary in B, K ∩ Q = g [K ∩ Q ∩ ω2]. As Q is 28 THOMAS GILTON AND JOHN KRUEGER

∗ ∗ elementary in B, K ∩ Q = g [K ∩ Q ∩ ω2] is in Q. It follows that for all K ∈ A , K ∩ Q ∈ Q. Let B := A∗ ∩ Q. We will show that for any C ≤ B in P ∩ Q, A∗ ∪ C is S- adequate, which ﬁnishes the proof. So let C ≤ B be in P ∩ Q. By the previous claim, for all K ∈ A∗, K ∩ Q ∈ A∗ ∩ Q. And A∗ ∩ Q ⊆ C ⊆ Q. By Proposition 1.35, A∗ ∪ C is adequate. To prove that A∗ ∪ C is S-adequate, let L ∈ C and N ∈ A∗, and we will show that RL(N) ∪ RN (L) ⊆ S.

By Lemma 2.9(2), RN (L) ⊆ RN∩Q(L). Since L and N ∩Q are in C, RN∩Q(L) ⊆ S. Hence RN (L) ⊆ S. Let ζ ∈ RL(N). Then by Lemma 2.9(1), either ζ ∈ RL(N ∩Q), or ζ = min((N ∩ ω2) \ β). In the ﬁrst case, since L and N ∩ Q are in C and C is S-adequate, it follows that ζ ∈ S. Assume the second case. Then since β ∈ RM∩Q(M), and M, ∗ M ∩Q, and N are in A , by Lemma 4.1 we have that ζ is in RM∩Q(M), RM∩Q(N), ∗ or RM (N). Since A is S-adequate, it follows that ζ ∈ S.

Corollary 4.7. The forcing poset P is ω2-strongly proper on a stationary set. Therefore P preserves ω2 and has the ω2-covering property. Proof. Immediate from Lemma 4.6. Proposition 4.8. The forcing poset P adds a club subset of S ∪ cof(ω). Proof. Let C˙ be a P-name for the set [ {RM (N): ∃A ∈ G,˙ M, N ∈ A}. ˙ It follows easily from Lemma 4.6 that P forces that C is cofinal in ω2. We claim that P forces that lim(C˙ ) ⊆ S ∪ cof(ω), which completes the proof. Let β < ω2, and assume that A is a condition which forces that β is a limit point of C˙ . We will prove that β is in S ∪ cof(ω). If β has cofinality ω, then we are done. So assume that cf(β) = ω1. We will show that β ∈ S. Fix N ∈ X0 such that A and β are in N. Then A ∪ {N} is an S-adequate set, and thus is in P. Since A ∪ {N} ≤ A, A ∪ {N} forces that β is a limit point of C˙ with uncountable cofinality. Hence we can fix B ≤ A ∪ {N}, K and M in B, and γ ∈ RK (M) such that sup(N ∩ β) < γ < β.

Since β ∈ N, we have that β = min((N ∩ ω2) \ γ). By Lemma 4.1,

β ∈ RK (M) ∪ RK (N) ∪ RM (N). As B is S-adequate, it follows that β ∈ S.

We remark that it is not necessary to assume that κ is ω2. If κ > ω2, we can ﬁx any stationary set S ⊆ κ ∩ cof(>ω), and then the forcing poset P deﬁned above will add a club subset of S ∪ cof(ω), and collapse κ to become ω2 by Proposition 3.12. MITCHELL’S THEOREM REVISITED 29

§5. S~-obedient side conditions

We now generalize the idea of an S-adequate set to the case when we have a sequence S~ of sets, instead of a single set S. For the remainder of this section ﬁx a sequence S~ = hSη : η < λi, where each Sη is a subset of κ ∩ cof(>ω). + Deﬁnition 5.1. A set P ∈ Y0 is S~-strong if for all τ ∈ P ∩ κ , P ∩ κ ∈ Sτ .

Note that if P is S~-strong, then cf(P ∩ κ) > ω, since P ∩ κ ∈ S0 ⊆ κ ∩ cof(>ω). For the next two definitions, we fix a class Y ⊆ Y0. The definitions of S~-adequate and S~-obedient are made relative to the class Y. Definition 5.2. Let A be an adequate set. We say that A is S~-adequate if for all M and N in A and ζ ∈ RM (N): + (1) for all τ ∈ M ∩ N ∩ κ , ζ ∈ Sτ ; (2) if P ∈ M ∩ Y is S~-strong and sup(N ∩ ζ) < P ∩ κ < ζ, then for all + τ ∈ N ∩ P ∩ κ , ζ ∈ Sτ . Definition 5.3. A pair (A, B) is an S~-obedient side condition if: (1) A is an S~-adequate set;

(2) B is a ﬁnite set of S~-strong models in Y0; (3) for all M ∈ A and P ∈ B, if ζ = min((M ∩ κ) \ (P ∩ κ)), then for all + τ ∈ P ∩ M ∩ κ , ζ ∈ Sτ . The next two lemmas show that we can add certain models to an S~-obedient side condition and preserve S~-obedience. Lemma 5.4. Let (A, B) be an S~-obedient side condition.

(1) If N ∈ X0 and (A, B) ∈ N, then ({N}, ∅) and (A ∪ {N},B) are S~-obedient side conditions. (2) If P ∈ Y0 is S~-strong and (A, B) ∈ P , then (∅, {P }) and (A, B ∪ {P }) are S~-obedient side conditions. Proof. (2) is trivial. (1) The fact that ({N}, ∅) is an S~-obedient side condition is easy. The set A∪{N} is S~-adequate because for all M ∈ A, M ∩βM,N = M ∩κ is in N, and RM (N) and RN (M) are empty. If P ∈ B, then min((N∩κ)\(P ∩κ)) = P ∩κ. + ~ And if τ ∈ P ∩ N ∩ κ , then P ∩ κ ∈ Sτ since P is S-strong. Lemma 5.5. Let (A, B) be an S~-obedient side condition. (1) Let M and N be in A, and suppose that M < N. Then (A ∪ {M ∩ N},B) is an S~-obedient side condition. (2) Let M ∈ A and P ∈ B. Then (A ∪ {M ∩ P },B) is an S~-obedient side condition. (3) Suppose that P and Q are in B and P ∩ κ < Q ∩ κ. Then (A, B ∪ {P ∩ Q}) is an S~-obedient side condition. Proof. (1) The set A ∪ {M ∩ N} is adequate by Proposition 1.25. To show that A ∪ {M ∩ N} is S~-adequate, it suﬃces to show that for all K ∈ A, {K,M ∩ N} is S~-adequate. So let K ∈ A be given. + Let ζ ∈ RK (M ∩N). Then ζ ∈ RK (M) by Lemma 2.3. Let τ ∈ K∩(M ∩N)∩κ , and we will show that ζ ∈ Sτ . Then τ ∈ K ∩ M, which implies that ζ ∈ Sτ since ζ ∈ RK (M). 30 THOMAS GILTON AND JOHN KRUEGER

Suppose that P ∈ K ∩ Y is S~-strong and sup(M ∩ N ∩ ζ) < P ∩ κ < ζ. Let + τ ∈ (M ∩ N) ∩ P ∩ κ , and we will show that ζ ∈ Sτ . Since ζ ∈ M ∩ N ∩ κ, ζ < βM,N . And since M ∩ N ∩ κ = M ∩ βM,N is an initial segment of M ∩ κ, sup(M ∩ N ∩ ζ) = sup(M ∩ ζ). So sup(M ∩ ζ) < P ∩ κ < ζ. Since ζ ∈ RK (M), it follows that ζ ∈ Sτ . Let ζ ∈ RM∩N (K). Then by Lemma 2.3, either (i) ζ ∈ RM (K) or (ii) ζ ∈ + RN (K). Consider τ ∈ K ∩ (M ∩ N) ∩ κ , and we will show that ζ ∈ Sτ . Then τ ∈ K ∩ M, so in case (i), ζ ∈ Sτ . Also τ ∈ K ∩ N, so in case (ii), ζ ∈ Sτ . Suppose that P ∈ (M ∩N)∩Y is S~-strong and sup(K ∩ζ) < P ∩κ < ζ. Let τ ∈ K ∩P ∩κ+, and we will show that ζ ∈ Sτ . Then P ∈ M ∩ Y, so in case (i), ζ ∈ Sτ . And P ∈ N ∩ Y, so in case (ii), ζ ∈ Sτ . This completes the proof that A ∪ {M ∩ N} is S~-adequate. Let Q ∈ B, and suppose that ξ = min((M ∩ N ∩ κ) \ (Q ∩ κ)). Since M < N, M ∩ N ∩ κ = M ∩ βM,N , which is an initial segment of M ∩ κ. Hence ξ = + min((M ∩ κ) \ (Q ∩ κ)). Let τ ∈ (M ∩ N) ∩ Q ∩ κ , and we will show that ξ ∈ Sτ . Then τ ∈ M ∩ Q, so ξ ∈ Sτ since (A, B) is S~-obedient.

(2) Since P is S~-strong, cf(P ∩ κ) > ω. So clearly sup(M ∩ P ∩ κ) < P ∩ κ. It follows that A ∪ {M ∩ P } is adequate by Proposition 1.33. To show that A ∪ {M ∩ P } is S~-adequate, let K ∈ A be given, and we will show that {K,M ∩ P } is S~-adequate. + Let ζ ∈ RK (M ∩P ). Then ζ ∈ RK (M) by Lemma 2.8. Let τ ∈ K ∩(M ∩P )∩κ , and we will show that ζ ∈ Sτ . Then τ ∈ K ∩ M implies that ζ ∈ Sτ . Suppose that Q ∈ K ∩ Y is S~-strong and sup(M ∩ P ∩ ζ) < Q ∩ κ < ζ. Let + τ ∈ Q ∩ (M ∩ P ) ∩ κ , and we will show that ζ ∈ Sτ . Since ζ ∈ P ∩ κ and P ∩ κ is an ordinal, sup(M ∩ P ∩ ζ) = sup(M ∩ ζ). So sup(M ∩ ζ) < Q ∩ κ < ζ. Since ζ ∈ RK (M), Q ∈ K ∩ Y, and τ ∈ M ∩ Q, it follows that ζ ∈ Sτ . Let ζ ∈ RM∩P (K). Then by Lemma 2.8, either (i) ζ ∈ RM (K) or (ii) ζ = min((K ∩ κ) \ (P ∩ κ)). + Let τ ∈ (M ∩ P ) ∩ K ∩ κ , and we will show that ζ ∈ Sτ . In case (i), τ ∈ M ∩ K implies that ζ ∈ Sτ since A is S~-adequate. In case (ii), τ ∈ K ∩ P implies that ζ ∈ Sτ since (A, B) is S~-obedient. Suppose that Q ∈ (M ∩ P ) ∩ Y is S~-strong and sup(K ∩ ζ) < Q ∩ κ < ζ. Let + τ ∈ K ∩ Q ∩ κ , and we will show that ζ ∈ Sτ . In case (i), Q ∈ M ∩ Y implies that ζ ∈ Sτ since A is S~-adequate. In case (ii), since τ ∈ Q and Q ∈ P , τ ∈ P . So ζ ∈ Sτ since (A, B) is S~-obedient. This completes the proof that A ∪ {M ∩ P } is S~-adequate. Let Q ∈ B, and suppose that ζ = min((M ∩ P ∩ κ) \ (Q ∩ κ)). Let τ ∈ + (M ∩ P ) ∩ Q ∩ κ , and we will show that ζ ∈ Sτ . Since P ∩ κ ∈ κ, M ∩ P ∩ κ is an initial segment of M ∩ κ. Hence ζ = min((M ∩ κ) \ (Q ∩ κ)). Since τ ∈ M ∩ Q, it follows that ζ ∈ Sτ since (A, B) is S~-obedient.

(3) Note that since P ∩ κ < Q ∩ κ, P ∩ Q ∩ κ = P ∩ κ. To show that P ∩ Q is S~-strong, let τ ∈ P ∩ Q ∩ κ+. Then τ ∈ P . Since P is S~-strong, P ∩ Q ∩ κ = P ∩ κ ∈ Sτ . Let M ∈ A, and suppose that ζ = min((M ∩ κ) \ (P ∩ Q ∩ κ)). Then ζ = + min((M ∩ κ) \ (P ∩ κ)). Let τ ∈ M ∩ (P ∩ Q) ∩ κ . Then τ ∈ M ∩ P , so ζ ∈ Sτ . MITCHELL’S THEOREM REVISITED 31

We conclude the section with an easy lemma which will be used frequently for checking that certain models are S~-strong.

Lemma 5.6. Suppose that N ∈ X0 ∪ Y0, Q ∈ N ∩ Y0, and P ∈ Y0. Suppose + that P is S~-strong and N ≺ (H(λ), ∈, Y0, S~). Assume that Q ∩ N ∩ κ ⊆ P and P ∩ κ = Q ∩ κ. Then Q is S~-strong. Proof. Since Q ∈ N, it suﬃces to show that N models that Q is S~-strong. So let τ ∈ Q ∩ N ∩ κ+. Since Q ∩ N ∩ κ+ ⊆ P , τ ∈ P . As P is S~-strong, Q ∩ κ = P ∩ κ ∈ Sτ .

§6. The approximation property and factorization

We brieﬂy discuss the approximation property, and state the theorem on factor- ing a generic extension which we will use in the proof of Mitchell’s theorem in Part III. Let (W1,W2) be a pair of transitive class models of ZFC such that W1 ⊆ W2. We say that the pair (W1,W2) satisﬁes the ω1-approximation property if, whenever X ∈ W2 is a set of ordinals such that a ∩ X ∈ W1 whenever a ∈ W1 is countable in W1, then the set X itself is in W1. The approximation property is due to Hamkins [5], and is similar to properties studied in Mitchell’s construction of a model with no Aronszajn trees on ω2 [11]. It plays a crucial role in the original proof of Mitchell’s theorem on the approachability ideal, as well as in the proof presented in Part III. We will use the following easy consequence of the approximation property.

Lemma 6.1. Suppose that (W1,W2) satisﬁes the ω1-approximation property. As- sume that c is a set of ordinals of order type ω1 in W2 such that for all β < sup(c), c ∩ β ∈ W1. Then c ∈ W1.

Proof. To show that c ∈ W1, it suﬃces to show that for any set a ∈ W1 which is countable in W1, a∩c ∈ W1. So let a ∈ W1 be countable in W1. Then a is countable in W2. Since c has order type ω1, there is β < sup(c) such that a ∩ c ⊆ c ∩ β. By the assumption on c, c ∩ β ∈ W1. Since c ∩ β and a are in W1, a ∩ c = a ∩ (c ∩ β) is in W1.

In the original proof of Mitchell’s theorem, being able to factor a generic extension in a way which satisﬁes the approximation property relied on what was called tidy strongly generic conditions (see Lemma 2.22 of [12]). However, the strongly generic conditions used in the present paper are not tidy. Therefore we need a diﬀerent factorization theorem which is applicable in the present context; such a theorem was provided by Cox-Krueger [2].

Let us recall the property ∗ introduced in [2].

Deﬁnition 6.2. Let P1 be a suborder of a forcing poset P2, where P2 has greatest lower bounds. We say that P1 satisﬁes property ∗(P1, P2) if for all p ∈ P1 and q, r ∈ P2, if p, q, and r are pairwise compatible in P2, then p is compatible in P2 with q ∧ r. 32 THOMAS GILTON AND JOHN KRUEGER

Note that if a forcing poset Q satisﬁes property ∗(Q, Q), then for any suborder P of Q, ∗(P, Q).

Notation 6.3. Let Q be a forcing poset. If q ∈ Q and K is a subset of Q, let (Q/q)/K denote the forcing poset consisting of conditions s ∈ Q such that s ≤ q, and s is compatible in Q with all conditions in K. The following result appears as Theorem 4.3 in [2]. Theorem 6.4 (Factorization theorem). Let Q be a forcing poset with greatest lower bounds satisfying ∗(Q, Q), χ a regular cardinal with λQ ≤ χ, and N ≺ (H(χ), ∈ , Q). Suppose that there are stationarily many models in Pω1 (H(χ)) which have universal strongly generic conditions. Assume that q is a universal strongly N- generic condition. Then for any V -generic filter G on Q which contains q, V [G] = V [G ∩ N][H], where G ∩ N is a V -generic filter on Q ∩ N, H is a V [G ∩ N]-generic filter on (Q/q)/(G ∩ N), and the pair (V [G ∩ N],V [G]) satisfies the ω1-approximation property. This theorem will be used in the final argument of the proof of Mitchell’s theorem in Section 16. It is interesting to note that not all intermediate extensions of a strongly proper forcing extension satisfy the ω1-approximation property; see Section 5 of [2] for a counterexample.

Part 2. Advanced side condition methods

3 §7. Mitchell’s use of κ κ + For the remainder of the paper we will assume κ and 2 = κ . Also we let the cardinal λ from Part I equal κ+. Since 2κ = κ+, H(κ+) has size κ+. Notation 7.1. Let f ∗ denote a bijection from κ+ to H(κ+).

+ Notation 7.2. Fix a sequence C~ = hCα : α < κ , α limiti satisfying that for all limit α < κ+:

(1) Cα is a club subset of α with ot(Cα) ≤ κ; in particular, if cf(α) < κ then ot(Cα) < κ; (2) if β ∈ lim(Cα), then Cβ = Cα ∩ β; (3) if α is a limit of limit ordinals, then every ordinal in Cα is a limit ordinal;

(4) if α = α0 + ω for a limit ordinal α0, then α0 ∈ lim(Cα), and hence Cα0 = Cα ∩ α0. Properties (1) and (2) embody the standard deﬁnition of a square sequence. It is easy to modify a square sequence to also satisfy properties (3) and (4). For example, start by replacing each ordinal in Cα with the greatest limit ordinal less than or equal to it. The details are left to the reader. + Notation 7.3. For each limit ordinal α < κ and β < ot(Cα), let cα,β denote the β-th member of Cα, that is, the unique γ in Cα such that ot(Cα ∩ γ) = β.

3Almost all of the arguments in this section and the next are due to Mitchell, but adopted to the present context. MITCHELL’S THEOREM REVISITED 33

+ Notation 7.4. Fix a sequence A~ = hAη,β : η < κ , β < κi satisfying the following properties: + (1) for each η < κ , {Aη,β : β < κ} is an increasing and continuous sequence of sets with union equal to η; (2) Aη+1,β = Aη,β ∪ {η}; + (3) for all η < κ and β < κ, |Aη,β| ≤ |β| · ω; (4) if ξ ∈ lim(Cη) ∪ Aη,β ∪ lim(Aη,β), then Aξ,β = Aη,β ∩ ξ; (5) there exists a function c∗ : κ → κ such that for all η < κ+ and β < κ, ∗ ot(Aη,β) < c (β); (6) if β < ot(Cη), then Aη,β ⊆ cη,β and lim(Cη) ∩ cη,β ⊆ Aη,β; (7) if γ ∈ η \ Cη, then γ ∈ Aη,β iﬀ

γ ∈ Amin(Cη \γ),β and min(Cη \ γ) ∈ Aη,β;

(8) if ξ ∈ lim(Aη,β) ∩ η and ot(Cξ) < β, then ξ ∈ Aη,β. Properties (1), (2), and (3) describe a typical kind of ﬁltration of each ordinal η < κ+. The coherence property (4) is one of the most often used facts in the paper. It gives suﬃcient conditions for coherence to hold between Aξ,β and Aη,β, where ξ < η. If ξ is a limit point of Cη, then

∀β < κ Aξ,β = Aη,β ∩ ξ.

And if β < κ and ξ is either in Aη,β, or a limit point of Aη,β, then

Aξ,β = Aη,β ∩ ξ. We recommend that the reader memorize this important fact before proceeding. Property (5) follows immediately from property (3) in the case when κ is weakly inaccessible, by letting c∗(β) = β+. This property is only used in one lemma in the paper, namely Lemma 8.6. Likewise, properties (6), (7), and (8) are technical facts about A~ which are only used in Lemmas 8.10 and 8.11, and in several places in Section 12. There is no harm in the reader forgetting about properties (5)–(8) for now, and just looking back at them later in the rare places that they are used. Theorem 7.5 (Mitchell [12]). Assume that κ is weakly inaccessible and C~ is a sequence as in Notation 7.2. Then there exists a sequence A~ as described in Notation 7.4. Mitchell constructs the sequence A~ using the square sequence C~ in a careful way. The only place where the weak inaccessibility of κ is used is to derive property (5) from property (3), as mentioned above. If κ is weakly inaccessible in an inner model + W + V ~ W which satisﬁes κ, and (κ ) = (κ ) , then the sequence A constructed in W still satisﬁes properties (1)–(8) in V by upwards absoluteness. For example, if V is + obtained from W by collapsing κ to become ω2 while preserving κ , then there is a sequence A~ as above in V . The construction of A~ appears in [12, Section 3.1]. We do not repeat it here because it is technical and not helpful for understanding the other material in our paper. Notation 7.6. Let A denote some expansion of the structure + ∗ ∗ ∗ ∗ ~ ~ ∗ (H(κ ), ∈, E, κ, T , π ,C , Λ, Y0, f , C, A, c ). 34 THOMAS GILTON AND JOHN KRUEGER

Note that A is an expansion of the structure described in Notations 1.9 and 1.10. Thus any elementary substructure of A is also an elementary substructure of that structure. + ∗ Notation 7.7. Let X denote the set of M in Pω1 (H(κ )) such that M ∩ κ ∈ T , M ≺ A, and lim(Csup(M)) ∩ M is coﬁnal in sup(M). + Notation 7.8. Let Y denote the set of P in Pκ(H(κ )) such that P ∩ κ ∈ κ, P ≺ A, and lim(Csup(P )) ∩ P is coﬁnal in sup(P ).

Note that X ⊆ X0 and Y ⊆ Y0, where X0 and Y0 were defined in Notations 1.9 and 1.10.4 Observe that by elementarity, for any M ∈ X and P ∈ Y, M = f ∗[M ∩ κ+] and P = f ∗[P ∩ κ+]. In particular, if M and N are in X ∪ Y and M ∩ κ+ ∈ N, then by elementarity, M ∈ N. As a result of the presence of the well-ordering E, the structure A described in Notation 7.6 has definable Skolem functions. Let hτn : n < ωi be a complete list of definable Skolem terms for A. For any set a ⊆ H(κ+), let Sk(a) denote the closure of a under the Skolem terms. 0 For n < ω and m equal to the arity of τn, we define a partial function τn : + m + 0 (κ ) → κ by letting τn(α0, . . . , αm−1) = τn(α0, . . . , αm−1), provided that this 0 is an ordinal, and otherwise is undefined. Note that τn is also definable in A. Notation 7.9. Let H∗ :(κ+)<ω → κ+ be a function such that any elementary substructure of A is closed under H∗, and whenever a ⊆ κ+ is closed under H∗, + ∗ 0 then Sk(a) ∩ κ = a. In addition, a is closed under H iff a is closed under τn for all n < ω. The existence of such a function H∗ is proved by standard arguments. Note that if a is a set of ordinals closed under H∗, then Sk(a) = f ∗[a]. In particular, if M ∈ X ∪ Y and a ∈ M is a set of ordinals which is closed under H∗, then by elementarity, Sk(a) ∈ M. The next simple lemma will prove very useful throughout the paper. Lemma 7.10. Suppose that N ∈ X ∪ Y, a is a set of ordinals in N, and for some set b which is closed under H∗, N ∩ a = N ∩ b. Then a is closed under H∗. 0 Proof. It suffices to show that a is closed under τn for all n < ω. Fix n < ω, 0 0 and let k be the arity of τn. Since a ∈ N and τn is definable in A, it suffices to 0 show that N models that a is closed under τn. Let α0, . . . , αk−1 ∈ N ∩ a. Then ∗ α0, . . . , αk−1 ∈ N ∩ b. Since N and b are both closed under H , they are closed 0 0 0 under τn. So τn(α0, . . . , αk−1) ∈ N ∩ b. Since N ∩ b ⊆ a, τn(α0, . . . , αk−1) ∈ a. In the remainder of this section, we will provide a thorough analysis of the models in X and Y. The following notation will be useful. + + Notation 7.11. Let N ⊆ H(κ ) be a set and γ ∈ κ ∩ sup(N). Let γN denote the ordinal min((N ∩ κ+) \ γ). Recall that if a is a set of ordinals, then cl(a) denotes the set a ∪ lim(a).

4 The requirement that lim(Csup(X)) ∩ X is cofinal in sup(X) appears in Mitchell’s definition of a model ([12, Definition 3.14]). We do not, however, assume that ot(Csup(X)) ∈/ X, as in his definition. MITCHELL’S THEOREM REVISITED 35

Lemma 7.12. Let N ∈ X ∪ Y and η ∈ cl(N), and suppose that η < sup(N). Then either η ∈ N or η ∈ lim(CηN ). Hence:

(1) Cη = CηN ∩ η;

(2) Aη,ξ = AηN ,ξ ∩ η for all ξ < κ;

(3) N ∩ AηN ,ξ = N ∩ Aη,ξ for all ξ < κ.

Proof. If η ∈ N, then ηN = η, and (1), (2), and (3) are trivial.

Suppose that η < ηN , and we will show that η ∈ lim(CηN ). Let γ < η. Since η ∈ cl(N) \ N, η ∈ lim(N). So we can ﬁx σ ∈ (N ∩ η) \ γ. Now ηN ∈ N and σ ∈ N ∩ ηN , so by elementarity there is δ ∈ CηN ∩ N larger than σ. Then

δ ∈ N ∩ ηN ⊆ η. So γ < δ < η and δ ∈ CηN . This proves that η ∈ lim(CηN ). (1) follows from the deﬁnition of a square sequence, and (2) follows from Notation 7.4(4). For (3), since N ∩ ηN = N ∩ η, it follows that for all ξ < κ,

N ∩ AηN ,ξ = N ∩ AηN ,ξ ∩ η = N ∩ Aη,ξ.

Lemma 7.13. Let N ∈ X ∪ Y, and suppose that η ∈ cl(N) \ N. Then lim(Cη) ∩ N is cofinal in η. Proof. Note that η ∈ lim(N). If η = sup(N), then the statement of the lemma follows from the definitions of X and Y. Otherwise by Lemma 7.12, Cη = CηN ∩ η. Let γ < η. Since η ∈ lim(N), we can fix σ ∈ N ∩ η larger than γ. As η < ηN , ηN has uncountable cofinality. So certainly lim(CηN ) is cofinal in ηN . By elementarity, we can find δ ∈ lim(CηN ) ∩ ηN ∩ N which is larger than σ. Then δ < η. Since

Cη = CηN ∩ η, it follows that δ ∈ lim(Cη). Thus γ ≤ δ and δ ∈ lim(Cη) ∩ N. The next lemma is standard.

+ + Lemma 7.14. Suppose that P ∈ Pκ(H(κ )), P ≺ (H(κ ), ∈), P ∩ κ ∈ κ, and cf(P ∩ κ) > ω. Assume that γ is a limit point of P ∩ κ+ below sup(P ), and cf(γ) < cf(P ∩ κ). Then γ ∈ P .

Proof. Suppose for a contradiction that γ∈ / P . Then γP is in P and γ < γP . By elementarity, we can fix an increasing and cofinal function f : cf(γP ) → γP which + is in P . Since γP < κ , either cf(γP ) < κ or cf(γP ) = κ. In the first case, cf(γP ) ∈ P ∩κ ∈ κ implies that cf(γP ) ⊆ P . By elementarity, f[cf(γP )] ⊆ P ∩γP ⊆ γ, which is impossible since f[cf(γP )] is cofinal in γP and γ < γP . Therefore cf(γP ) = κ. By elementarity, f P ∩ κ is cofinal in P ∩ γP , and hence is cofinal in γ. But then γ has cofinality equal to cf(P ∩ κ), which contradicts our assumption on γ.

+ Lemma 7.15. Let P ∈ Pκ(H(κ )) with P ∩ κ ∈ κ and P ≺ A. If cf(P ∩ κ) > ω and cf(sup(P )) > ω, then P ∈ Y.

Proof. Let σ := sup(P ). By the definition of Y, it suffices to show that lim(Cσ)∩P is cofinal in σ. So let ξ < σ. Since sup(P ) = σ has uncountable cofinality, there exists a sequence hγn : n < ωi increasing and bounded below σ such that ξ < γ0, γn ∈ P if n is even, and γn ∈ Cσ if n is odd. Let γ be the supremum of this sequence. Then γ is a limit point of P which is strictly below sup(P ) with cofinality ω. Since cf(P ∩ κ) > ω, cf(γ) < cf(P ∩ κ). So γ ∈ P by Lemma 7.14. On the other hand, γ is a limit point of Cσ. So ξ < γ and γ ∈ lim(Cσ) ∩ P . 36 THOMAS GILTON AND JOHN KRUEGER

Lemma 7.16. Let M and N be in X ∪ Y, and assume that {M,N} is adequate in the case that M and N are in X . Then M ∩ N ∈ X ∪ Y. Specifically: (1) if M ∈ X and N ∈ X ∪ Y, then M ∩ N ∈ X ; (2) if M ∈ Y and N ∈ Y, then M ∩ N ∈ Y. Proof. Obviously M ∩ N is an elementary substructure of A. If M ∈ X , then M ∩ N ∈ X0 by Lemma 1.23 and the comment after Notation 1.10. Hence M ∩ N ∩ κ ∈ T ∗. And if M and N are in Y, then M ∩ N ∩ κ = min{M ∩ κ, N ∩ κ} ∈ κ. Let α := sup(M ∩ N). It remains to show that lim(Cα) ∩ (M ∩ N) is cofinal in α. First we claim that lim(Cα)∩M and lim(Cα)∩N are cofinal in α. Since M ∩N ∩ κ+ is closed under successors, it does not have a maximum element, and therefore α + is a limit point of M ∩N ∩κ . As α ∈ lim(M), if α∈ / M then lim(Cα)∩M is cofinal in α by Lemma 7.13, and similarly with N. So if α is neither in M nor N, then the claim is proved. Assume that α is in one of them. Since α = sup(M ∩ N), α cannot be in both in M and N. Without loss of generality, assume that α ∈ N \M. Then lim(Cα) ∩ M is cofinal in α as just observed, and so in particular, lim(Cα) is cofinal in α. By the elementarity of N, and since α ∈ N and also α is a limit point of N, easily lim(Cα) ∩ N is cofinal in α. This completes the proof of the claim. 0 + To show that lim(Cα) ∩ (M ∩ N) is cofinal in α, let γ < α. Fix γ ∈ M ∩ N ∩ κ 0 0 with γ < γ . Let σ = min(lim(Cα)\γ ). We claim that σ ∈ M ∩N, which completes the proof. Since lim(Cα)∩M is cofinal in α, we can fix η ∈ lim(Cα)∩M with σ < η. 0 0 As η ∈ lim(Cα), Cη = Cα ∩ η. Therefore σ = min(lim(Cη) \ γ ). Since η and γ are in M, σ ∈ M by elementarity. The same argument shows that σ ∈ N. We now introduce the idea of a simple model.5 These are the models for which there exist strongly generic conditions. To motivate the definition, we prove a bound on ot(Csup(N)).

Lemma 7.17. Let N ∈ X ∪ Y. If η ∈ lim(N), then ot(Cη) ∈ cl(N ∩ κ). In particular, ot(Csup(N)) ≤ sup(N ∩ κ). Proof. Since η ∈ lim(N) and |N| < κ, it follows that cf(η) < κ. If η ∈ N, then ot(Cη) ∈ N ∩ κ by elementarity. Assume that η is not in N. Then η ∈ cl(N) \ N. By Lemma 7.13, lim(Cη) ∩ N is coﬁnal in η. We claim that ot(Cη) is a limit point of N ∩ κ. Let γ < ot(Cη). Then we can ﬁnd δ ∈ lim(Cη) ∩ N such that

γ < ot(Cη ∩ δ) = ot(Cδ) < ot(Cη).

Since δ ∈ N, ot(Cδ) ∈ N ∩ot(Cη). So γ < ot(Cδ) < ot(Cη) and ot(Cδ) ∈ N ∩κ.

Deﬁnition 7.18. Let N ∈ X ∪ Y. We say that N is simple if ot(Csup(N)) = sup(N ∩ κ). We prove next that there exist stationarily many simple models in X .

Lemma 7.19. Let N ∈ X ∪ Y and δ := sup(N). Then for all ξ ∈ N ∩ ot(Cδ), cδ,ξ ∈ N.

Proof. Let ξ ∈ N ∩ ot(Cδ). Then ξ < ot(Cδ). As lim(Cδ) ∩ N is coﬁnal δ, we can ﬁx η ∈ lim(Cδ) ∩ N such that ξ < ot(Cδ ∩ η) = ot(Cη). Hence cδ,ξ = cη,ξ. Since η and ξ are in N, by elementarity, cη,ξ is in N.

5This is the same idea as described in [12, Section 3.3]. MITCHELL’S THEOREM REVISITED 37

Proposition 7.20. The collection of models in X which are simple is stationary + in Pω1 (H(κ )). Proof. Let F : H(κ+)<ω → H(κ+), and we will find a simple model in X which is closed under F . Fix X of size κ such that X ≺ A, X is closed under F , and + θ := X ∩ κ has cofinality κ. Let c : κ → θ be the function c(ξ) = cθ,ξ for all ξ < κ. ∗ Since T is stationary in Pω1 (κ) and X ≺ A, we can find M ∈ Pω1 (X) which is closed under F such that ∗ M ∩ κ ∈ T ,M ≺ A, and M ≺ (X, ∈,Cθ, c). Let δ := sup(M). We claim that M is in X and is simple. To show that M ∈ X , it suffices to show that lim(Cδ) ∩ M is cofinal in δ. By elementarity, clearly δ ∈ lim(Cθ). Hence Cδ = Cθ ∩ δ. As M is closed under c, for all ξ ∈ M ∩ κ, + c(ξ) = cθ,ξ ∈ M ∩ κ ⊆ δ.

Thus c(ξ) ∈ Cθ ∩ δ = Cδ. So c(ξ) = cθ,ξ = cδ,ξ. It follows by elementarity that {c(ξ): ξ ∈ M ∩ κ} is increasing and coﬁnal in M ∩ δ. In particular, the set {c(ξ): ξ ∈ M ∩ κ, ξ limit} witnesses that lim(Cδ) ∩ M is coﬁnal δ. It remains to show that M is simple, which means that ot(Cδ) = sup(M ∩κ). We know that ot(Cδ) ≤ sup(M ∩ κ) by Lemma 7.17. Suppose for a contradiction that ot(Cδ) < sup(M∩κ). Fix β ∈ (M∩κ)\ot(Cδ). Then c(β) ∈ M by elementarity. But c(β) = cθ,β = cδ,β, as previously observed, which is absurd since ot(Cδ) ≤ β. Regarding the stationarity of simple models in Y, see Lemma 8.3 and Proposition 14.2.

Next we will show that a model M in X ∪ Y is determined by sup(M ∩ κ) and sup(M). + Notation 7.21. Consider η < κ and β < κ. For γ < ot(Aη,β), let aη,β,γ be equal to the γ-th element of Aη,β. Deﬁne πη : κ × κ → η by letting πη(γ, β) = aη,β,γ if γ < ot(Aη,β), and 0 otherwise.

Note that πη is a surjection of κ × κ onto η. Also if ξ ∈ Aη,β, then ξ = πη(ot(Aη,β ∩ ξ), β). Observe that πη is deﬁnable in the structure A. Lemma 7.22. Let η < κ+ and β < κ. Suppose that

δ ∈ lim(Cη) ∪ Aη,β ∪ lim(Aη,β) and γ < ot(Aδ,β). Then aη,β,γ = aδ,β,γ . So πη(γ, β) = πδ(γ, β).

Proof. By Notation 7.4(4), Aδ,β = Aη,β ∩ δ, so clearly aη,β,γ = aδ,β,γ . Lemma 7.23. Let N ∈ X ∪ Y. Then + N ∩ κ = {πsup(N)(γ, β): γ, β ∈ N ∩ κ}. Proof. Let η := sup(N). Suppose that γ and β are in N ∩ κ, and we will show that πη(γ, β) ∈ N. This is obvious if πη(γ, β) = 0. So assume that γ < ot(Aη,β) and πη(γ, β) = aη,β,γ . Since N ∈ X ∪ Y, lim(Cη) ∩ N is coﬁnal in η. As aη,β,γ < η, we can ﬁx δ ∈ lim(Cη) ∩ N such that aη,β,γ < δ. Since Aδ,β = Aη,β ∩ δ, clearly γ < ot(Aδ,β). By Lemma 7.22, πη(γ, β) = πδ(γ, β). As δ, γ, and β are in N, πδ(γ, β) ∈ N by elementarity. 38 THOMAS GILTON AND JOHN KRUEGER

Conversely, let ξ ∈ N ∩ κ+ be given, and we will find γ and β in N ∩ κ such that πη(γ, β) = ξ. Since lim(Cη) ∩ N is cofinal in η, we can fix δ ∈ lim(Cη) ∩ N such that ξ < δ. Then ξ and δ are in N. By elementarity, there is β ∈ N ∩ κ such that ξ ∈ Aδ,β. Let γ := ot(Aδ,β ∩ ξ). Since δ, β, and ξ are in N, γ ∈ N. As noted after Notation 7.21, aδ,β,γ = πδ(γ, β) = ξ. Since δ ∈ lim(Cη), by Lemma 7.22, πη(γ, β) = πδ(γ, β) = ξ. Lemma 7.24. Let M and N be in X ∪ Y, and suppose that M ∩ κ and sup(M) are in N. Then M ∈ N. Proof. Since M = f ∗[M ∩κ+], by elementarity it suffices to show that M ∩κ+ ∈ N. Let η := sup(M). Then + M ∩ κ = {πη(γ, β): γ, β ∈ M ∩ κ} + by Lemma 7.23. Since η and M ∩ κ are in N, M ∩ κ ∈ N by elementarity.

The next topic we consider is the set Asup(M),sup(M∩κ), where M ∈ X ∪ Y.

+ Lemma 7.25. Let M ∈ X ∪ Y. Then M ∩ κ ⊆ Asup(M),sup(M∩κ).

+ Proof. Let ξ ∈ M ∩ κ . Since lim(Csup(M)) ∩ M is cofinal in sup(M), we can fix σ ∈ lim(Csup(M)) ∩ M which is strictly greater than ξ. By elementarity, we can fix β ∈ M ∩ κ such that ξ ∈ Aσ,β. Since σ ∈ lim(Csup(M)), we have that Aσ,β = Asup(M),β ∩ σ. Hence ξ ∈ Asup(M),β. As β ∈ M ∩ κ, β < sup(M ∩ κ). Therefore Asup(M),β ⊆ Asup(M),sup(M∩κ). Hence ξ ∈ Asup(M),sup(M∩κ).

+ Lemma 7.26. Let Q ∈ Y. Then Q ∩ κ = Asup(Q),Q∩κ.

+ Proof. By Lemma 7.25, we have that Q ∩ κ ⊆ Asup(Q),Q∩κ. Conversely, let ξ ∈ Asup(Q),Q∩κ, and we will show that ξ ∈ Q. Since lim(Csup(Q)) ∩ Q is cofinal in sup(Q), we can fix σ ∈ lim(Csup(Q)) ∩ Q which is strictly greater than ξ. As σ ∈ lim(Csup(Q)), Aσ,Q∩κ = Asup(Q),Q∩κ ∩ σ. Therefore ξ ∈ Aσ,Q∩κ. Since Q ∩ κ is a limit ordinal, we can fix β < Q ∩ κ such that ξ ∈ Aσ,β. Then σ and β are in Q, and hence Aσ,β ∈ Q. Since |Aσ,β| < κ by Notation 7.4(3), Aσ,β ⊆ Q. Therefore ξ ∈ Q.

Lemma 7.27. Let Q ∈ Y and η ∈ cl(Q). Then Q ∩ η = Aη,Q∩κ.

+ Proof. By Lemma 7.26, Q ∩ κ = Asup(Q),Q∩κ. Since η ∈ cl(Q),

η ∈ Asup(Q),Q∩κ ∪ lim(Asup(Q),Q∩κ). By Notation 7.4(4),

Aη,Q∩κ = Asup(Q),Q∩κ ∩ η = Q ∩ η.

Lemma 7.28. Suppose that P1 and P2 are in Y.

(1) If P1 ∩ κ ≤ P2 ∩ κ and η ∈ cl(P1) ∩ cl(P2), then P1 ∩ η ⊆ P2 ∩ η. (2) If P1 ∩ κ < P2 ∩ κ and η ∈ P1 ∩ P2 ∩ cof(κ), then P1 ∩ η ∈ P2. In particular, sup(P1 ∩ η) ∈ P2 ∩ η. MITCHELL’S THEOREM REVISITED 39

Proof. By Lemma 7.27, under the assumptions of either (1) or (2), we have that

P1 ∩ η = Aη,P1∩κ, and P2 ∩ η = Aη,P2∩κ.

(1) If P1 ∩κ ≤ P2 ∩κ, then clearly Aη,P1∩κ ⊆ Aη,P2∩κ. Therefore P1 ∩η ⊆ P2 ∩η. (2) If P1 ∩ κ < P2 ∩ κ, then P1 ∩ κ ∈ P2. So P1 ∩ κ and η are in P2, and hence

Aη,P1∩κ = P1 ∩ η is in P2 by elementarity. Since η has coﬁnality κ, sup(P1 ∩ η) < η. So sup(P1 ∩ η) ∈ P2 ∩ η.

Lemma 7.29. Let M ∈ X . Let A := Asup(M),sup(M∩κ). Then A is closed under H∗, A ∩ κ = sup(M ∩ κ), and sup(A) = sup(M). Proof. To show that A is closed under H∗, it suﬃces to show that for each n < ω, 0 A is closed under τn. At the same time, we will show that sup(M ∩ κ) ⊆ A. So ﬁx 0 n < ω, and let k be the arity of τn. Let α0, . . . , αk−1 ∈ A and β < sup(M ∩ κ), and 0 we will show that τn(α0, . . . , αk−1) and β are in A. Fix η0 ∈ lim(Csup(M))∩M such that α0, . . . , αk−1 and βM are strictly less than η0. Then A ∩ η0 = Aη0,sup(M∩κ).

So α0, . . . , αk−1 are in Aη0,sup(M∩κ). Also βM ∈ M ∩ κ ⊆ A by Lemma 7.25, so also βM ∈ Aη0,sup(M∩κ). As sup(M ∩ κ) is a limit ordinal, we can ﬁx an inﬁnite

γ ∈ M ∩ κ such that α0, . . . , αk−1 and βM are in Aη0,γ . By the elementarity of M, we can fix η1 ∈ M strictly greater than η0 such that 0 η1 is closed under τn. Fix η2 ∈ lim(Csup(M)) ∩ M with η1 < η2. Since η0 < η1 0 0 and η1 is closed under τn, for all γ0, . . . , γk−1 in Aη0,γ , τn(γ0, . . . , γk−1) < η1 < η2. Define a function h : Ak → κ by letting h(γ , . . . , γ ) be the least ordinal ξ < κ η0,γ 0 k−1 0 such that τn(γ0, . . . , γk−1) and all ordinals below βM are in Aη2,ξ. Since η0, γ, η2, and βM are in M, by elementarity h is in M. Now the domain of h has size |Ak | ≤ |γ| < κ. So there exists a minimal η0,γ ξ < κ such that h[Ak ] ⊆ ξ. By elementarity, ξ ∈ M ∩ κ. In particular, δ := η0,γ 0 h(α0, . . . , αk−1) is less than ξ. That means τn(α0, . . . , αk−1) and β are in Aη2,δ ⊆ 0 Aη2,ξ. Since η2 ∈ lim(Csup(M)), Aη2,ξ = Asup(M),ξ ∩ η2. So τn(α0, . . . , αk−1) and β are in Asup(M),ξ ⊆ Asup(M),sup(M∩κ) = A. This completes the proof that A is closed 0 ∗ under τn for all n < ω and sup(M ∩ κ) ⊆ A. It follows that A is closed under H . Now A is the union of sets of the form Aδ,β, where δ ∈ lim(Csup(M)) ∩ M and β ∈ M ∩ κ, and each such set is in M. Thus each such set Aδ,β satisfies that sup(Aδ,β ∩ κ) < sup(M ∩ κ). It follows that sup(A ∩ κ) ≤ sup(M ∩ κ). But we just proved that sup(M ∩ κ) ⊆ A, and therefore sup(M ∩ κ) = A ∩ κ. By the definition of A, obviously A ⊆ sup(M). And since M ∩ κ+ ⊆ A by Lemma 7.25, sup(A) = sup(M).

We conclude this section with two technical lemmas which will be useful later. Lemma 7.30. Let N ∈ X , a ∈ N, and τ ∈ N ∩ κ+. Suppose that cf(τ) > ω. If a ∩ [sup(N ∩ τ), τ) 6= ∅, then τ is a limit point of a. Proof. If τ is not a limit point of a, then sup(a ∩ τ) < τ. Since a and τ are in N, sup(a ∩ τ) ∈ N ∩ τ by elementarity. Hence sup(a ∩ τ) < sup(N ∩ τ), which contradicts the assumption that a ∩ [sup(N ∩ τ), τ) 6= ∅. Lemma 7.31. Let N ∈ X . Suppose that η ∈ N ∩ κ+ and β ∈ N ∩ κ. Let

ξ ∈ Aη,β \ N. Then ξ ∈ AξN ,β. 40 THOMAS GILTON AND JOHN KRUEGER

Proof. Note that since η ∈ N and ξ < η, ξN exists. Since ξ∈ / N, ξ < ξN . It follows that cf(ξN ) > ω, since otherwise N ∩ ξN would be coﬁnal in ξN by elementarity. Also sup(N ∩ ξN ) ≤ ξ. Since Aη,β ∈ N and ξ ∈ Aη,β ∩ [sup(N ∩ ξN ), ξN ), it follows that ξN is a limit point of Aη,β by Lemma 7.30. So AξN ,β = Aη,β ∩ ξN . Since

ξ ∈ Aη,β ∩ ξN , ξ ∈ AξN ,β.

§8. Interaction of models past κ

The method of adequate sets, which we dealt with in Part I, handles the interaction of countable elementary substructures below κ. In this section we will show how the coherent ﬁltration system A~ from Section 7 can be used to control the interaction of models between κ and κ+.

Notation 8.1. For M and N in X ∪ Y, let αM,N denote the ordinal sup(M ∩ N). As we discussed in Section 1 in the paragraph after Propostion 1.29, if M < N, in general it does not necessarily follow that M ∩N ∈ N. The next lemma describes a situation in which this implication does hold. Lemma 8.2. Let M and N be in X ∪ Y, where N is simple. Suppose that: (1) M and N are in X and M < N, or (2) M and N are in Y and M ∩ κ < N ∩ κ, or (3) M ∈ X , N ∈ Y, and sup(M ∩ N ∩ κ) < N ∩ κ.

Then M ∩ N ∈ N. In particular, αM,N ∈ N. Proof. By Lemma 1.30, M ∩N ∩κ ∈ N. Therefore by elementarity, cl(M ∩N ∩κ) ∈ N. We claim that cl(M ∩ N ∩ κ) ⊆ N ∩ κ. If M ∩ N is countable, then so is cl(M ∩ N ∩ κ). Since cl(M ∩ N ∩ κ) ∈ N, it follows that cl(M ∩ N ∩ κ) ⊆ N ∩ κ. If M ∩ N is uncountable, then M and N are both in Y. So M ∩ κ < N ∩ κ by (2), and hence M ∩ N ∩ κ = M ∩ κ. Therefore cl(M ∩ N ∩ κ) = (M ∩ κ) ∪ {M ∩ κ}, which is a subset of N ∩ κ. Next we claim that + (N ∩ κ ) \ αM,N 6= ∅. + Suppose for a contradiction that (N ∩ κ ) \ αM,N = ∅, which means that sup(N) = αM,N . Since N is simple, it follows that

ot(CαM,N ) = sup(N ∩ κ).

But ot(CαM,N ) ∈ cl(M ∩ N ∩ κ) by Lemma 7.17. By the ﬁrst claim, it follows that ot(CαM,N ) ∈ N ∩ κ, which contradicts that ot(CαM,N ) = sup(N ∩ κ). + Let α := min((N ∩ κ ) \ αM,N ). By Lemma 7.23, + M ∩ N ∩ κ = {παM,N (γ, β): γ, β ∈ M ∩ N ∩ κ}.

We claim that for all γ, β ∈ M ∩ N ∩ κ, παM,N (γ, β) = πα(γ, β). This is immediate if α = αM,N , so assume that αM,N ∈/ N. Then by Lemma 7.12, αM,N ∈ lim(Cα). Fix γ and β in M ∩ N ∩ κ.

First, assume that ot(Aα,β) ≤ γ. Then πα(γ, β) = 0. Since AαM,N ,β = Aα,β ∩

αM,N , clearly ot(AαM,N ,β) ≤ ot(Aα,β) ≤ γ. So παM,N (γ, β) = 0. Secondly, assume that γ < ot(Aα,β), so that πα(γ, β) = aα,β,γ . Since α, γ, and β are in N, aα,β,γ ∈ MITCHELL’S THEOREM REVISITED 41

N ∩ α ⊆ αM,N . As AαM,N ,β = Aα,β ∩ αM,N , clearly γ < ot(AαM,N ,β). By Lemma

7.22, πα(γ, β) = παM,N (γ, β). It follows that + M ∩ N ∩ κ = {πα(γ, β): γ, β ∈ M ∩ N ∩ κ}. Since α and M ∩ N ∩ κ are in N, so is M ∩ N ∩ κ+ by elementarity. Hence by ∗ + elementarity, M ∩ N = f [M ∩ N ∩ κ ] ∈ N, and sup(M ∩ N) = αM,N ∈ N. Lemma 8.3. Let P ∈ Y, and assume that cf(sup(P )) = P ∩ κ. Then P is simple.

Proof. Since Csup(P ) is coﬁnal in sup(P ),

P ∩ κ = cf(sup(P )) ≤ ot(Csup(P )). On the other hand, as sup(P ) ∈ lim(P ), Lemma 7.17 implies that

ot(Csup(P )) ∈ cl(P ∩ κ) = (P ∩ κ) ∪ {P ∩ κ}.

Hence ot(Csup(P )) ≤ P ∩ κ. Therefore P ∩ κ = ot(Csup(P )), and P is simple. Lemma 8.4. Let P ∈ Y, and assume that cf(sup(P )) = P ∩ κ. If M ∈ X , then M ∩ P ∈ P . If Q ∈ Y and Q ∩ κ < P ∩ κ, then Q ∩ P ∈ P . Proof. By Lemma 8.3, P is simple. Since cf(sup(P )) = P ∩ κ, P ∩ κ is a regular cardinal. Obviously ω < P ∩ κ, so P ∩ κ is a regular uncountable cardinal. If M ∈ X , then sup(M ∩P ∩κ) < P ∩κ since sup(M ∩P ∩κ) has countable coﬁnality. By Lemma 8.2, we are done. Lemma 8.5. Let M ∈ X and N ∈ X ∪ Y, where {M,N} is adequate if N ∈ X , and sup(M ∩ N ∩ κ) < N ∩ κ if N ∈ Y. Then

lim(M) ∩ lim(N) ⊆ αM,N + 1.

Proof. Let η ∈ lim(M) ∩ lim(N), and we will show that η ≤ αM,N . By Lemma 7.17, ot(Cη) ∈ cl(M ∩ κ) ∩ cl(N ∩ κ). Since η is a limit point of M and M is countable, η has cofinality ω. Therefore ot(Cη) has cofinality ω. We claim that ot(Cη) is a limit point of M ∩ κ. If ot(Cη) ∈ M, then since ot(Cη) has countable cofinality, easily M ∩ot(Cη) is cofinal in ot(Cη) by elementarity. Hence ot(Cη) is a limit point of M ∩ κ. Otherwise if ot(Cη) ∈/ M, then since ot(Cη) ∈ cl(M ∩ κ), it follows immediately that ot(Cη) is in lim(M ∩ κ). Next we claim that ot(Cη) ∈ cl(M ∩ N ∩ κ). First, assume that N ∈ X . Then by Lemma 1.20,

ot(Cη) ∈ cl(M ∩ κ) ∩ cl(N ∩ κ) = cl(M ∩ N ∩ κ). Secondly, assume that N ∈ Y. Now

ot(Cη) ∈ cl(N ∩ κ) = (N ∩ κ) ∪ {N ∩ κ}.

So ot(Cη) ≤ N ∩ κ. Since sup(M ∩ N ∩ κ) < N ∩ κ and ot(Cη) is a limit point of M ∩ κ, it cannot be the case that ot(Cη) = N ∩ κ. Therefore ot(Cη) < N ∩ κ. By Lemma 1.31, cl(M ∩ N ∩ κ) = cl(M ∩ κ) ∩ cl(N ∩ κ) ∩ (N ∩ κ).

Since ot(Cη) is in the set on the right, it is in cl(M ∩ N ∩ κ). 42 THOMAS GILTON AND JOHN KRUEGER

Now we claim that ot(Cη) ∈ lim(M ∩ N ∩ κ).

As ot(Cη) ∈ cl(M ∩N ∩κ), either ot(Cη) ∈ M ∩N ∩κ, or ot(Cη) ∈ lim(M ∩N ∩κ). In the latter case, we are done. In the former case, since ot(Cη) has cofinality ω, by the elementarity of M ∩ N, clearly M ∩ N ∩ κ is cofinal in ot(Cη), so again ot(Cη) ∈ lim(M ∩ N ∩ κ). Finally, we are ready to prove that η ≤ αM,N . Suppose for a contradiction that αM,N < η. Since ot(Cη) is a limit point of M ∩N ∩κ, we can fix γ ∈ M ∩N ∩ot(Cη) such that αM,N < cη,γ . We claim that cη,γ ∈ M ∩ N, which is a contradiction + since M ∩ N ∩ κ ⊆ αM,N . If η ∈ M, then obviously cη,γ ∈ M by elementarity. Otherwise η ∈ cl(M) \ M. By Lemma 7.13, lim(Cη) ∩ M is cofinal in η. So we can fix δ ∈ lim(Cη) ∩ M such that cη,γ < δ. Then clearly cη,γ = cδ,γ , which is in M by elementarity. The proof that cη,γ ∈ N is the same.

We now turn to address the following general issue. Suppose that M and N are in X ∪ Y, and P ∈ N ∩ Y. Under what circumstances can we conclude that an ordinal in M ∩ P is in N, or is in some canonically described member of N? The next lemma is the most frequently used result on this topic. Lemma 8.6. Let M and N be in X , where M ≤ N. Suppose that η ∈ N ∩ κ+ and β < sup(M ∩ N ∩ κ).

Then Aη,β ∩ M ⊆ N. Therefore

Aη,sup(M∩N∩κ) ∩ M ⊆ N.

Proof. Let ξ ∈ Aη,β ∩ M, and we will show that ξ ∈ N. Since β < sup(M ∩ N ∩ κ), we can ﬁx γ ∈ M ∩N ∩κ greater than β. Then ξ ∈ Aη,γ . So ξ = πη(ot(Aη,γ ∩ξ), γ), as noted in the comments after Notation 7.21. Since ξ ∈ Aη,γ , Aξ,γ = Aη,γ ∩ ξ. Therefore ot(Aη,γ ∩ ξ) = ot(Aξ,γ ). Hence ξ = πη(ot(Aξ,γ ), γ). Since ξ and γ are in M, so is ot(Aξ,γ ). Since γ ∈ M ∩ N ∩ κ, by elementarity c∗(γ) ∈ M ∩ N ∩ κ. By Notation 7.4(5), since M ≤ N, we have that ∗ ot(Aξ,γ ) ∈ M ∩ c (γ) ⊆ M ∩ N ∩ κ ⊆ N.

So ot(Aξ,γ ) ∈ N ∩ κ. Hence η, γ, and ot(Aξ,γ ) are in N, which implies that πη(ot(Aξ,γ ), γ) = ξ is in N. To show that Aη,sup(M∩N∩κ) ∩ M ⊆ N, let τ ∈ Aη,sup(M∩N∩κ) ∩ M. Since sup(M ∩ N ∩ κ) is a limit ordinal, there is β < sup(M ∩ N ∩ κ) such that τ ∈ Aη,β. By what we just proved, Aη,β ∩ M ⊆ N. So τ ∈ N. Lemma 8.7. Let M and N be in X , where M ≤ N. Let Q ∈ N ∩ Y with Q ∩ κ < sup(M ∩ N ∩ κ). Then Q ∩ M ∩ κ+ ⊆ N.

+ + Proof. By Lemma 7.26, Q ∩ κ = Asup(Q),Q∩κ. By elementarity, sup(Q) ∈ N ∩ κ , and by assumption, Q ∩ κ < sup(M ∩ N ∩ κ). By Lemma 8.6, + Q ∩ M ∩ κ = Asup(Q),Q∩κ ∩ M ⊆ N. Lemma 8.8. Let M and N be in X . MITCHELL’S THEOREM REVISITED 43

(1) If M ≤ N, then

AαM,N ,sup(M∩N∩κ) ∩ M ⊆ N. (2) If M ∼ N, then

AαM,N ,sup(M∩N∩κ) ∩ M = AαM,N ,sup(M∩N∩κ) ∩ N. Proof. Note that (2) follows from (1). To prove (1), assume that M ≤ N, and let ξ ∈ AαM,N ,sup(M∩N∩κ) ∩ M. We will show that ξ ∈ N. Fix β ∈ M ∩ N ∩ κ such that ξ ∈ AαM,N ,β. As M ∩ N ∈ X and sup(M ∩ N) = αM,N , it follows that lim(CαM,N ) ∩ (M ∩ N) is coﬁnal in αM,N . So we can ﬁx δ ∈ lim(CαM,N ) ∩ (M ∩ N) which is strictly larger than ξ. Then Aδ,β = AαM,N ,β ∩ δ, and hence ξ ∈ Aδ,β. Since δ ∈ N, β < sup(M ∩ N ∩ κ), and M ≤ N, it follows that Aδ,β ∩ M ⊆ N by Lemma 8.6. So ξ ∈ N. Lemma 8.9. Let M ∈ X and N ∈ X ∪ Y. Then + M ∩ N ∩ κ ⊆ AαM,N ,sup(M∩N∩κ).

Proof. Since M ∩ N ∈ X and sup(M ∩ N) = αM,N , the statement follows immediately from Lemma 7.25. Lemma 8.10. Let M ∈ X and N ∈ X ∪ Y. Let Q ∈ M ∩ Y, and suppose that sup(M ∩ N ∩ κ) ≤ Q ∩ κ. Then

Q ∩ N ∩ αM,N ⊆ AαM,N ,Q∩κ.

Proof. Let ξ ∈ Q ∩ N ∩ αM,N , and we will show that ξ ∈ AαM,N ,Q∩κ. First assume + that ξ ∈ M. Then ξ ∈ M ∩ N ∩ κ , so by Lemma 8.9, ξ ∈ AαM,N ,sup(M∩N∩κ).

Since sup(M ∩ N ∩ κ) ≤ Q ∩ κ, it follows that ξ ∈ AαM,N ,Q∩κ. + Assume that ξ is not in M. Then ξ ∈ Q ∩ κ = Asup(Q),Q∩κ, where sup(Q) and Q ∩ κ are in M, and ξ∈ / M. By Lemma 7.31,

ξ ∈ AξM ,Q∩κ. We claim that + ∀ν ∈ M ∩ N ∩ κ (ξM < ν =⇒ ξ ∈ Aν,Q∩κ). We will prove the claim by induction. So let ν ∈ M ∩ N ∩ κ+ be strictly greater 0 than ξM , and assume that the claim holds for all ν ∈ M ∩ N ∩ ν.

Case 1: ν = ν0 + 1 is a successor ordinal. Since ν ∈ M ∩ N, ν0 ∈ M ∩ N by elementarity. If ξM < ν0, then by the inductive hypothesis, ξ ∈ Aν0,Q∩κ. So

ξ ∈ Aν0,Q∩κ ∪ {ν0} = Aν,Q∩κ.

If ν0 = ξM , then

ξ ∈ AξM ,Q∩κ = Aν0,Q∩κ ⊆ Aν,Q∩κ.

Case 2: ν is a limit ordinal and ξM ∈ lim(Cν ). Then

AξM ,Q∩κ = Aν,Q∩κ ∩ ξM .

Since ξ ∈ AξM ,Q∩κ, it follows that ξ ∈ Aν,Q∩κ, and we are done.

0 Case 3: ν is a limit ordinal and ξM is not in lim(Cν ). Let ν := min(Cν \ ξM ), 00 0 0 00 and let ν := sup(Cν ∩ ν ). Since ν = min(Cν \ ξM ), clearly ν = sup(Cν ∩ ξM ). 00 As ξM is not a limit point of Cν , ν < ξM . 44 THOMAS GILTON AND JOHN KRUEGER

0 0 We claim that ν ∈ M ∩ N. Since ν and ξM are in M, ν = min(Cν \ ξM ) is 0 00 0 in M by elementarity. And as ν and ν are in M, ν = sup(Cν ∩ ν ) is in M by 00 elementarity. But ν < ξM and ξM is the least ordinal in M with ξ ≤ ξM . It 00 0 00 follows that ν < ξ. So ν = min(Cν \ (ν + 1)) = min(Cν \ ξ). Since ν and ξ are in N, so is ν0 by elementarity. 0 Next we claim that ξ ∈ Aν0,Q∩κ. This is immediate if ξM = ν , so assume that 0 0 0 + ξM < ν . Then ξM < ν < ν and ν ∈ M ∩ N ∩ κ , which imply by the inductive hypothesis that ξ ∈ Aν0,Q∩κ. 0 0 Let β be the least ordinal in κ such that ν ∈ Aν,β. Since ν and ν are in M ∩ N, it follows that β ∈ M ∩ N ∩ κ by elementarity. As β < sup(M ∩ N ∩ κ) ≤ Q ∩ κ, 0 00 0 we have that ν ∈ Aν,Q∩κ. And since ν = sup(Cν ∩ ν ) and 00 0 ν < ξ < ξM ≤ ν ,

ξ∈ / Cν . So 0 ξ ∈ ν \ Cν , ν = min(Cν \ ξ) ∈ Aν,Q∩κ, and ξ ∈ Amin(Cν \ξ),Q∩κ.

By Notation 7.4(7), ξ ∈ Aν,Q∩κ, which completes the proof of the claim.

Since M ∩N ∈ X and sup(M ∩N) = αM,N , it follows that lim(CαM,N )∩(M ∩N) is coﬁnal in αM,N . Since ξ < αM,N by assumption and αM,N is a limit point of

M, we have that ξM < αM,N . So we can ﬁx ν ∈ lim(CαM,N ) ∩ (M ∩ N) which is strictly greater than ξM . By the claim, ξ ∈ Aν,Q∩κ. Since ν ∈ lim(CαM,N ),

Aν,Q∩κ = AαM,N ,Q∩κ ∩ ν. Therefore ξ ∈ AαM,N ,Q∩κ. Lemma 8.11. Let M ∈ X and N ∈ X ∪ Y. Suppose that M < N in the case that N ∈ X . Let P ∈ M ∩ Y, and suppose that P ∩ κ ∈ M ∩ N ∩ κ and P ∩ αM,N is bounded below αM,N . Deﬁne

σ := sup(P ∩ AαM,N ,sup(M∩N∩κ)). Then σ satisﬁes: (1) σ ∈ M ∩ N ∩ κ+; (2) P ∩ σ = Aσ,P ∩κ; + (3) P ∩ (M ∩ N) ∩ κ = Aσ,P ∩κ ∩ (M ∩ N); (4) N ∩ P ∩ αM,N ⊆ Aσ,P ∩κ.

Proof. Let α := αM,N and δ := sup(M ∩ N ∩ κ). Note that by Lemma 7.29, + Sk(Aα,δ) ∩ κ = Aα,δ,Aα,δ ∩ κ = δ, and sup(Aα,δ) = α.

Since P and Aα,δ are closed under successors, P ∩ Aα,δ has no maximal element. Note that since P ∩ αM,N is bounded below αM,N , we have that σ < α. We claim that σ satisﬁes (1)–(4). Observe that since σ is a limit point of P , it follows that P ∩ σ = Aσ,P ∩κ by Lemma 7.27, which proves (2). (3) We prove that + P ∩ (M ∩ N) ∩ κ = Aσ,P ∩κ ∩ (M ∩ N). + Let γ ∈ P ∩ (M ∩ N) ∩ κ , and we will show that γ ∈ Aσ,P ∩κ. Since γ ∈ M ∩ N, by Lemma 8.9 we have that γ ∈ Aα,δ. So γ ∈ P ∩Aα,δ ⊆ σ. Hence γ ∈ P ∩σ = Aσ,P ∩κ. Conversely, Aσ,P ∩κ ∩ (M ∩ N) ⊆ Aσ,P ∩κ = P ∩ σ ⊆ P. MITCHELL’S THEOREM REVISITED 45

(1,4) It remains to show that σ ∈ M ∩N and N ∩P ∩αM,N ⊆ Aσ,P ∩κ. We claim that

σ = sup(Asup(P ∩α),P ∩κ ∩ Aα,δ). As P and α are closed under successors, P ∩α has no maximal element. So sup(P ∩ α) is a limit point of P , which by Lemma 7.27 implies that

P ∩ α = P ∩ sup(P ∩ α) = Asup(P ∩α),P ∩κ. Therefore

Asup(P ∩α),P ∩κ ∩ Aα,δ = P ∩ α ∩ Aα,δ = P ∩ Aα,δ. Taking supremums of both sides yields the claim. Next, we claim that σ ∈ Aα,δ. As P ∩ α and Aα,δ are closed under successors and P ∩ α = Asup(P ∩α),P ∩κ, it follows that σ is a limit point of Asup(P ∩α),P ∩κ and a limit point of Aα,δ. By Notation 7.4(8) and the fact that σ ∈ lim(Aα,δ) ∩ α, to show that σ ∈ Aα,δ it suﬃces to show that ot(Cσ) < δ. Since P ∩σ = Aσ,P ∩κ and σ is a limit point of P , it follows that σ = sup(Aσ,P ∩κ). If P ∩ κ < ot(Cσ), then by Notation 7.4(6), it follows that Aσ,P ∩κ ⊆ cσ,P ∩κ < σ. But this contradicts that σ = sup(Aσ,P ∩κ). Hence ot(Cσ) ≤ P ∩ κ < δ, which completes the proof of the claim that σ ∈ Aα,δ. Fix η ∈ lim(Cα) ∩ (M ∩ N) such that σ < η. Then Aη,δ = Aα,δ ∩ η. Therefore σ ∈ Aη,δ. Since δ = sup(M ∩ N ∩ κ), we can ﬁx γ ∈ M ∩ N ∩ κ such that σ ∈ Aη,γ . Then η and γ are in M ∩ N. Let us show that

σ = max(Aη,γ ∩ lim(P )). 0 0 Suppose for a contradiction that σ ∈ Aη,γ ∩ lim(P ) and σ < σ . Since η ∈ lim(Cα) 0 and γ < δ, σ ∈ Aα,δ ∩ lim(P ). But then by Lemma 7.27, it follows that 0 0 P ∩ σ = Aσ0,P ∩κ ⊆ Aσ0,δ = Aα,δ ∩ σ . Since σ0 is a limit point of P , there is τ ∈ P ∩ σ0 strictly greater than σ. Then τ ∈ P ∩ Aα,δ, which contradicts that σ = sup(P ∩ Aα,δ). Now we prove that σ ∈ M ∩N. Since η, γ, and P are in M, and σ = max(Aη,γ ∩ lim(P )), it follows that σ ∈ M by elementarity. On the other hand, σ ∈ M ∩ Aη,γ , where η ∈ N and γ ∈ M ∩N ∩κ. So σ ∈ N by Lemma 8.6, in the case when M and N are in X . If N ∈ Y, then σ ∈ Aη,γ ∈ N implies that σ ∈ N, since |Aη,γ | < κ. This proves that σ ∈ M ∩ N. Now we claim that N ∩ P ∩ αM,N ⊆ σ. This completes the proof, for then

N ∩ P ∩ αM,N ⊆ P ∩ σ = Aσ,P ∩κ.

Suppose for a contradiction that π ∈ N ∩ P ∩ αM,N and σ ≤ π. Let π0 := + min((P ∩ κ ) \ σ), and note that π0 ≤ π. Since P and σ are in M, π0 is in M by elementarity. We claim that π0 is in N. This is immediate if π0 = π, so assume that π0 < π. Then π ∈ P implies that P ∩ π = Aπ,P ∩κ by Lemma 7.27. Since π and P ∩κ are in N, so is Aπ,P ∩κ = P ∩π. But π0 = min((P ∩π)\σ). Hence π0 ∈ N by elementarity. So π0 ∈ M ∩ N ∩ αM,N , and therefore π0 ∈ Aα,δ by Lemma 8.9. So π0 ∈ P ∩ Aα,δ. Since σ = sup(P ∩ Aα,δ) and P ∩ Aα,δ has no maximal element as previously observed, it follows that π0 < σ. But this contradicts that fact that σ ≤ π0. 46 THOMAS GILTON AND JOHN KRUEGER

So far in this section we have been mostly concerned about the interaction of models M and N below αM,N = sup(M ∩ N). We now turn to analyze what happens above αM,N . The next two lemmas state that for a simple model N, if a model does not bound N below κ, then it does not bound N above κ. Lemma 8.12. Let M and N be in X , where N is simple and {M,N} is adequate. + If RM (N) 6= ∅, then (N ∩ κ ) \ αM,N 6= ∅.

Proof. Since RM (N) is nonempty, βM,N ≤ sup(N ∩ κ). As αM,N is a limit point of M and a limit point of N, Lemma 7.17 implies that ot(CαM,N ) is in cl(M ∩ κ) ∩ + cl(N ∩ κ). Hence by Lemma 1.15, ot(CαM,N ) < βM,N . If (N ∩ κ ) \ αM,N is empty, then sup(N) = αM,N . Since N is simple, it follows that ot(CαM,N ) = sup(N ∩ κ). But then sup(N ∩ κ) < βM,N , which contradicts the first line above. Lemma 8.13. Let N ∈ X be simple and Q ∈ Y. If Q ∩ κ < sup(N ∩ κ), then sup(N ∩ Q) < sup(N). Proof. Let β := Q∩κ and η := sup(N ∩Q), and assume that β < sup(N ∩κ). Since N and Q are closed under successors, η is a limit point of N ∩ Q. In particular, η is a limit point of N. Suppose for a contradiction that sup(N) = η. Then since N is simple, ot(Cη) = sup(N ∩ κ). So β < ot(Cη). As N ∩ Q is in X by Lemma 7.16 and sup(N ∩ Q) = η, it follows that lim(Cη) ∩ (N ∩ Q) is cofinal in η. So we can fix δ ∈ lim(Cη) ∩ (N ∩ Q) such that β < ot(Cη ∩ δ) = ot(Cδ). But δ ∈ Q, and therefore by elementarity, ot(Cδ) ∈ Q ∩ κ = β. So ot(Cδ) < β, which is a contradiction.

We now introduce an analogue of remainder points for ordinals between αM,N and κ+. + Definition 8.14. Let M and N be in X ∪ Y. Define RN (M) as the set of ordinals η such that either: + (1) η = min((M ∩ κ ) \ αM,N ) and αM,N < η, or + + (2) η = min((M ∩ κ ) \ ξ), for some ξ ∈ (N ∩ κ ) \ αM,N . Lemma 8.15. Let M ∈ X and N ∈ X ∪ Y, where {M,N} is adequate if N ∈ X , and sup(M ∩ N ∩ κ) < N ∩ κ if N ∈ Y. Then: + (1) RN (M) is finite; + (2) if η ∈ RN (M), then cf(η) > ω; + + (3) suppose that η ∈ RN (M), η is not equal to min((M ∩κ )\αM,N ), and σ := + + + min((N ∩ κ ) \ sup(M ∩ η)); then σ ∈ RM (N) and η = min((M ∩ κ ) \ σ). + Proof. (1) If RN (M) is not finite, then the supremum of the first ω many members + of RN (M) is a limit point of M and a limit point of N. Hence this supremum is less + than or equal to αM,N by Lemma 8.5, which contradicts the definition of RN (M). + (2) is easy. (3) Note that σ exists, since otherwise η would not be in RN (M). + + Clearly η = min((M ∩ κ ) \ σ). We will show that σ ∈ RM (N). Since η is not + equal to min((M ∩ κ ) \ αM,N ), fix θ ∈ (M ∩ η) \ αM,N . Then αM,N ≤ θ < σ, + + and therefore αM,N < σ. If σ = min((N ∩ κ ) \ αM,N ), then σ ∈ RM (N) by definition. So assume not. Then we can fix ξ ∈ (N ∩ σ) \ αM,N . By definition, + + + ζ0 := min((M ∩ κ ) \ ξ) is in RN (M). Since ξ < σ = min((N ∩ κ ) \ sup(M ∩ η)) and ξ ∈ N, it follows that ξ < sup(M ∩ η). Therefore ζ0 < sup(M ∩ η) ≤ σ. MITCHELL’S THEOREM REVISITED 47

+ Let ζ1 be the largest member of RN (M) which is below σ. Then ζ1 exists since + RN (M)∩σ is ﬁnite and nonempty, as witnessed by ζ0. We claim that σ = min((N ∩ + + + κ ) \ ζ1), which proves that σ ∈ RM (N). Otherwise σ0 := min((N ∩ κ ) \ ζ1) is + strictly below σ. So σ0 < σ = min((N ∩ κ ) \ sup(M ∩ η)), which implies that + σ0 < sup(M ∩ η). But then min((M ∩ η) \ σ0) is in RN (M) ∩ σ, and is strictly larger than ζ1, which contradicts the maximality of ζ1. + + Lemma 8.16. Let M and N be in X ∪ Y. Then for all η ∈ RN (M) ∪ RM (N), η is closed under H∗.

+ Proof. First consider η = min((N ∩ κ ) \ αM,N ). Let n < ω and let k be the arity 0 0 of τn, and we will show that η is closed under τn. Since η ∈ N, by elementarity 0 it suﬃces to show that N models that η is closed under τn. Let α0, . . . , αk−1 ∈ N ∩ η. By the minimality of η, N ∩ η ⊆ αM,N , so α0, . . . , αk−1 < αM,N . By the + elementarity of M ∩N and since sup(M ∩N) = αM,N , there is some γ ∈ M ∩N ∩κ 0 such that α0, . . . , αk−1 are below γ and γ is closed under τn. Then 0 τn(α0, . . . , αk−1) < γ < αM,N ≤ η. + The same proof works for min((M ∩ κ ) \ αM,N ). + Now we prove the general statement by induction on ordinals in RN (M) ∪ + + + + RM (N). Suppose that η ∈ RN (M), and for all σ ∈ (RN (M) ∪ RM (N)) ∩ η, σ ∗ + is closed under H . If η = min((M ∩κ )\αM,N ), then we are done by the previous paragraph. Otherwise by Lemma 8.15(3), the ordinal σ := min((N ∩ κ+) \ sup(M ∩ η)) + + is in RM (N), and η = min((M ∩ κ ) \ σ). By the inductive hypothesis, σ is closed under H∗. 0 Let n < ω, and we will show that η is closed under τn. Let k be the arity of 0 τn. Since η is in M, by elementarity it suﬃces to show that M models that η is 0 closed under τn. Let α0, . . . , αk−1 ∈ M ∩ η. Then α0, . . . , αk−1 are strictly less than sup(M ∩ η) ≤ σ. Since σ is closed under H∗, 0 τn(α0, . . . , αk−1) < σ < η. + The same argument works for ordinals in RM (N).

§9. Canonical models

In this section we will introduce some models which are determined by canonical parameters which arise in the comparison of two models. Speciﬁcally, we consider a simple model N ∈ X and a model M in X ∪ Y which is not necessarily a member of N. The canonical models associated with M and N are models in N ∩ Y which reﬂect some information about M inside N. Canonical models will be used when amalgamating side conditions or forcing conditions over a simple model N; see Sections 13 and 15.6 The three types of canonical models are described in Notations 9.1, 9.3, and 9.13.

6The idea of a canonical model is new to this paper, and does not appear in Mitchell’s original proof [12]. 48 THOMAS GILTON AND JOHN KRUEGER

Notation 9.1. Let N ∈ X be simple and P ∈ Y, where P ∩ κ < sup(N ∩ κ). Let

β := P ∩ κ and η := sup(N ∩ P ). We let Q(N,P ) denote the set Sk(AηN ,βN ).

Note that ηN exists by Lemma 8.13. It is easy to check that if P ∈ N ∩ Y, then Q(N,P ) = P . Lemma 9.2. Let N ∈ X be simple and P ∈ Y, where P ∩ κ < sup(N ∩ κ). Let β := P ∩ κ and η := sup(N ∩ P ). Then Q := Q(N,P ) satisﬁes the following properties: (1) Q ∈ N ∩ Y; + (2) Q ∩ κ = βN , Q ∩ κ = AηN ,βN , and sup(Q) = ηN ; (3) N ∩ Q ∩ κ+ = N ∩ P ∩ κ+.

Proof. Since ηN and βN are in N, AηN ,βN and Q are in N by elementarity. As N ∩ P is closed under ordinal successors, η is a limit point of N ∩ P . Therefore + P ∩ η = Aη,β by Lemma 7.27. And since η is a limit point of N ∩ κ , by Lemma

7.12, Cη = CηN ∩ η, and for all ξ < κ,

Aη,ξ = AηN ,ξ ∩ η and N ∩ AηN ,ξ = N ∩ Aη,ξ. We claim that + N ∩ P ∩ κ = N ∩ AηN ,βN . + Let α ∈ N ∩ P ∩ κ , and we will show that α ∈ AηN ,βN . Then α < η by the deﬁnition of η. So

α ∈ P ∩ η = Aη,β = AηN ,β ∩ η.

Hence α ∈ AηN ,β. Since β ≤ βN , α ∈ AηN ,βN . Conversely, let α ∈ N ∩ AηN ,βN , and we will show that α ∈ P . Since ηN , βN , and α are in N and βN is a limit ordinal, by elementarity we can ﬁx ξ ∈ N ∩ βN such that α ∈ AηN ,ξ. Then

α ∈ N ∩ AηN ,ξ = N ∩ Aη,ξ. Since ξ ∈ N ∩ βN , ξ < β. So Aη,ξ ⊆ Aη,β = P ∩ η. Hence α ∈ P . + We have proven that N ∩ P ∩ κ = N ∩ AηN ,βN . By Lemma 7.10, it follows that ∗ + AηN ,βN is closed under H . In particular, Q ∩ κ = AηN ,βN . Therefore + + N ∩ Q ∩ κ = N ∩ AηN ,βN = N ∩ P ∩ κ , which proves (3). To show that Q ∩ κ = βN and sup(Q) = ηN , it suﬃces to prove that N models these statements. Let α ∈ N ∩ Q ∩ κ, and we will show that α < βN . Then

α ∈ N ∩ Q ∩ κ = N ∩ P ∩ κ ⊆ P ∩ κ = β ≤ βN .

So α < βN . Conversely, let α ∈ N ∩ βN , and we will show that α ∈ Q. Then α ∈ N ∩ βN ⊆ β. So α ∈ N ∩ β = N ∩ P ∩ κ = N ∩ Q ∩ κ. So indeed α ∈ Q. + Since Q ∩ κ = AηN ,βN , clearly sup(Q) ≤ ηN . To show that N models that sup(Q) = ηN , let ξ ∈ N ∩ ηN . Then ξ < η = sup(N ∩ P ). So we can ﬁx σ ∈ N ∩ P ∩ κ+ which is larger than ξ. Then σ ∈ N ∩ P ∩ κ+ = N ∩ Q ∩ κ+. So σ ∈ Q and ξ ≤ σ. Thus sup(Q) = ηN . This completes the proof of (2).

To show that Q ∈ Y, it suﬃces to prove that lim(CηN ) ∩ Q is coﬁnal in ηN . Again it will be enough to show that N models this statement. So let ξ ∈ N ∩ ηN . Then ξ < η = sup(N ∩ P ). Since N ∩ P ∈ X by Lemma 7.16 and sup(N ∩ P ) = η, MITCHELL’S THEOREM REVISITED 49

there is σ ∈ lim(Cη) ∩ (N ∩ P ) with ξ ≤ σ. Since Cη = CηN ∩ η, σ ∈ lim(CηN ). + + Also σ ∈ N ∩ P ∩ κ = N ∩ Q ∩ κ . So σ ∈ lim(CηN ) ∩ Q and ξ ≤ σ.

Notation 9.3. Let M and N be in X , where N is simple. Let ζ ∈ RM (N) and + η := min((N ∩ κ ) \ αM,N ). We let Q(N, M, ζ) denote the set Sk(Aη,ζ ). Note that η exists by Lemma 8.12. Lemma 9.4. Let M and N be in X , where N is simple. Suppose that N ≤ M and + ζ = min((N ∩ κ) \ βM,N ). Let η := min((N ∩ κ ) \ αM,N ). Then Q := Q(N, M, ζ) satisﬁes the following properties: (1) Q ∈ N ∩ Y; + (2) Q ∩ κ = ζ, Q ∩ κ = Aη,ζ , and sup(Q) = η; (3) N ∩ Q ∩ κ+ = M ∩ N ∩ κ+.

Proof. Since η and ζ are in N, Aη,ζ and Q are in N by elementarity. As αM,N is a limit point of N, by Lemma 7.12, CαM,N = Cη ∩ αM,N , and for all ξ < κ,

AαM,N ,ξ = Aη,ξ ∩ αM,N and N ∩ AαM,N ,ξ = N ∩ Aη,ξ. We claim that + N ∩ Aη,ζ = M ∩ N ∩ κ . + + Let α ∈ M ∩N ∩κ , and we will show that α ∈ Aη,ζ . By Lemma 8.9, M ∩N ∩κ ⊆

AαM,N ,sup(M∩N∩κ). Since ζ ∈ RM (N), sup(M ∩ N ∩ κ) < βM,N ≤ ζ. So + α ∈ M ∩ N ∩ κ ⊆ AαM,N ,sup(M∩N∩κ) ⊆ AαM,N ,ζ ⊆ Aη,ζ .

Hence α ∈ Aη,ζ . Conversely, let α ∈ N ∩ Aη,ζ , and we will show that α ∈ M. Since α, η, and ζ are in N and ζ is a limit ordinal, by elementarity we can ﬁx γ ∈ N ∩ ζ such that α ∈ Aη,γ . Since ζ = min((N ∩ κ) \ βM,N ) and N ≤ M, γ ∈ N ∩ βM,N = M ∩ N ∩ κ. So

α ∈ N ∩ Aη,γ = N ∩ AαM,N ,γ ⊆ N ∩ AαM,N ,sup(M∩N∩κ). By Lemma 8.8(1), α ∈ M. + We have proven that N ∩ Aη,ζ = M ∩ N ∩ κ . By Lemma 7.10, it follows that ∗ + Aη,ζ is closed under H . In particular, Q ∩ κ = Aη,ζ . Therefore + + N ∩ Q ∩ κ = N ∩ Aη,ζ = M ∩ N ∩ κ , which proves (3). To show that Q ∩ κ = ζ and sup(Q) = η, it suﬃces to show that N models these statements. Let γ ∈ N ∩ Q ∩ κ, and we will show that γ < ζ. Then γ ∈ N ∩ Q ∩ κ = M ∩ N ∩ κ, so

γ < sup(M ∩ N ∩ κ) < βM,N ≤ ζ. Conversely, let γ ∈ N ∩ ζ, and we will show that γ ∈ Q. Since ζ = min((N ∩ κ) \ βM,N ), γ ∈ N ∩ βM,N . As N ≤ M, N ∩ βM,N ⊆ M, so γ ∈ M. Hence γ ∈ M ∩ N ∩ κ+ ⊆ Q. + Since Q ∩ κ = Aη,ζ , obviously sup(Q) ≤ η. To show that N models that + sup(Q) = η, let ξ ∈ N ∩ η be given. Since η = min((N ∩ κ ) \ αM,N ), ξ < αM,N . + + As αM,N = sup(M ∩ N ∩ κ ), we can fix σ ∈ M ∩ N ∩ κ with ξ ≤ σ. Then σ ∈ M ∩ N ∩ κ+ ⊆ Q. So ξ ≤ σ and σ ∈ Q. This completes the proof of (2). To show that Q ∈ Y, it suffices to show that N models that lim(Cη)∩Q is cofinal in η. Let ξ ∈ N ∩η. Then ξ < αM,N . Since M ∩N is in X and sup(M ∩N) = αM,N , 50 THOMAS GILTON AND JOHN KRUEGER

we can ﬁx σ ∈ lim(CαM,N ) ∩ (M ∩ N) with ξ ≤ σ. But CαM,N = Cη ∩ αM,N , so + σ ∈ lim(Cη). Also σ ∈ M ∩ N ∩ κ ⊆ Q. So σ ∈ lim(Cη) ∩ Q and ξ ≤ σ. Notation 9.5. Let M and N be in X such that {M,N} is adequate. Let ζ ∈

RN (M). We let Q0(M, N, ζ) denote the set Sk(AαM,N ,ζ ). Lemma 9.6. Let M and N be in X such that {M,N} is adequate. Let η ∈ + 0 M ∩ N ∩ κ with κ ≤ η. Fix m < ω, and let k be the arity of τm. Deﬁne 0 fm,η : κ → κ by letting fm,η(β) be the least β < κ such that β ∈ Aη,β0 and 0 k η ∩ τm[Aη,β] ⊆ Aη,β0 .

Then for all σ ∈ RN (M) ∪ RM (N), σ is closed under fm,η. 0 k Note that since Aη,β has size less than κ by Notation 7.4(3), the set τm[Aη,β] also has size less than κ. So the deﬁnition of fm,η makes sense. Also note that fm,η is deﬁnable from η in A.

Proof. The proof is by induction on remainder points in RM (N) ∪ RN (M). For the base case, let σ be the first ordinal in RM (N)∪RN (M). Without loss of generality, assume that σ ∈ RN (M). Since η ∈ M and fm,η is definable from η in A, it suffices to show that M models that σ is closed under fm,η. So let β ∈ M ∩ σ, and we will show that fm,η(β) < σ. Since σ ∈ RN (M) and σ is the first remainder point, we have that M ≤ N and σ = min((M ∩ κ) \ βM,N ). As β ∈ M ∩ σ = M ∩ βM,N and M ≤ N, β ∈ M ∩ βM,N ⊆ N. So β ∈ N. Therefore η and β are both in M ∩ N. By elementarity, fm,η(β) ∈ M ∩ N ∩ κ. But M ∩ N ∩ κ = M ∩ βM,N ⊆ σ. So fm,η(β) < σ. Now suppose that ζ is a remainder point which is greater than the least remainder point, and assume that the lemma holds for all remainder points in RM (N)∪RN (M) which are below ζ. Without loss of generality, assume that ζ ∈ RN (M). Then by Lemma 2.2(3), σ := min((N ∩ κ) \ sup(M ∩ ζ)) is in RM (N), and ζ = min((M ∩ κ) \ σ). To show that M models that ζ is closed under fm,η, let β ∈ M ∩ ζ. Then β < sup(M ∩ ζ) < σ. By the inductive hypothesis, σ is closed under fm,η. So fm,η(β) < σ. Since σ < ζ, fm,η(β) < ζ.

Lemma 9.7. Let M and N be in X such that {M,N} is adequate. Let σ ∈ RN (M). Then Q0 := Q0(M, N, σ) satisﬁes the following properties:

(1) Q0 ∈ Y; + (2) Q0 ∩ κ = σ, Q0 ∩ κ = AαM,N ,σ, and sup(Q0) = αM,N ; + (3) M ∩ N ∩ κ ⊆ Q0.

Proof. Recall that Q0 = Q0(M, N, σ) = Sk(AαM,N ,σ). We begin by proving that ∗ 0 AαM,N ,σ is closed under H . Let m < ω, and let k be the arity of τm. Let 0 α0, . . . , αk−1 ∈ AαM,N ,σ, and we will show that τm(α0, . . . , αk−1) ∈ AαM,N ,σ. Since

σ is a limit ordinal, we can fix β < σ such that α0, . . . , αk−1 ∈ AαM,N ,β. By the elementarity of M ∩ N, fix δ ∈ M ∩ N ∩ κ+ strictly greater than 0 α0, . . . , αk−1 such that δ is closed under τm. Now fix η ∈ lim(CαM,N ) ∩ (M ∩ N) 0 strictly greater than δ and κ. Since δ is closed under τm, 0 τm(α0, . . . , αk−1) < δ < η.

As η ∈ lim(CαM,N ),

α0, . . . , αk−1 ∈ AαM,N ,β ∩ η = Aη,β. MITCHELL’S THEOREM REVISITED 51

So 0 0 k τm(α0, . . . , αk−1) ∈ η ∩ τm[Aη,β]. 0 As β < σ and σ ∈ RN (M), by Lemma 9.6 there is β < σ such that 0 k η ∩ τm[Aη,β] ⊆ Aη,β0 . Then 0 0 τm(α0, . . . , αk−1) ∈ Aη,β ⊆ Aη,σ = AαM,N ,σ ∩ η ⊆ AαM,N ,σ. ∗ + This proves that AαM,N ,σ is closed under H . In particular, Q0 ∩ κ = AαM,N ,σ. + Since M ∩ N ∩ κ ⊆ AαM,N ,sup(M∩N∩κ) by Lemma 8.9, and sup(M ∩ N ∩ κ) < σ, it + follows that M ∩N ∩κ ⊆ AαM,N ,σ ⊆ Q0. In particular, since sup(M ∩N) = αM,N , it follows that sup(Q0) = αM,N . It remains to show that Q0 ∈ Y and Q0 ∩κ = σ. For the first statement, once we know that Q0 ∩κ = σ, it will suffice to show that lim(CαM,N )∩Q0 is cofinal in αM,N .

But since M ∩ N ∈ X , lim(CαM,N ) ∩ (M ∩ N) is coﬁnal in sup(M ∩ N) = αM,N . + And as M ∩ N ∩ κ ⊆ Q0, it follows that lim(CαM,N ) ∩ Q0 is coﬁnal in αM,N . Now we prove that Q0 ∩ κ = σ. First we will show that Q0 ∩ κ ⊆ σ. More generally, we will prove by induction on remainder points that

∀ζ ∈ RM (N) ∪ RN (M),AαM,N ,ζ ∩ κ ⊆ ζ. Consider the ﬁrst remainder point ζ. Without loss of generality, assume that

ζ ∈ RN (M). Then M ≤ N and ζ = min((M ∩ κ) \ βM,N ). Let β ∈ AαM,N ,ζ ∩ κ, and we will show that β < ζ. Fix η ∈ lim(CαM,N ) ∩ (M ∩ N) with β < η. Then

β ∈ AαM,N ,ζ ∩ η = Aη,ζ . To show that β < ζ, it suffices to show that Aη,ζ ∩ κ ⊆ ζ. Since η and ζ are in M, it is enough to show that M models that Aη,ζ ∩ κ ⊆ ζ. 0 0 Let β ∈ M ∩Aη,ζ ∩κ, and we will show that β < ζ. Since ζ is a limit ordinal, by 0 elementarity we can fix γ ∈ M ∩ ζ with β ∈ Aη,γ . Since ζ = min((M ∩ κ) \ βM,N ) 0 and M ≤ N, γ ∈ M ∩ βM,N ⊆ N. As M ≤ N, β ∈ M ∩ Aη,γ , η ∈ N, and γ ∈ M ∩ N ∩ κ, it follows that β0 ∈ N by Lemma 8.6. Hence 0 β ∈ M ∩ N ∩ κ ⊆ βM,N ≤ ζ. So β0 < ζ. For the inductive step, let ζ be a remainder point which is not the first remainder point. Without loss of generality, assume that ζ ∈ RN (M). Then by Lemma 2.2(3), there is π ∈ RM (N) such that π = min((N ∩ κ) \ sup(M ∩ ζ)) and ζ = min((M ∩ κ) \ π). Let β ∈ AαM,N ,ζ ∩ κ, and we will show that β < ζ. Fix

η ∈ lim(CαM,N ) ∩ (M ∩ N) with β < η. Then

β ∈ AαM,N ,ζ ∩ η = Aη,ζ .

To show that β < ζ, it suffices to show that Aη,ζ ∩ κ ⊆ ζ. Since η and ζ are in M, by elementarity it suffices to show that M models that Aη,ζ ∩ κ ⊆ ζ. So let γ ∈ M ∩ Aη,ζ ∩ κ, and we will show that γ < ζ. Since ζ is a limit ordinal, by elementarity we can fix α ∈ M ∩ ζ such that γ ∈ Aη,α. Then α < sup(M ∩ ζ) < π. So γ ∈ Aη,α ⊆ Aη,π. Since η ∈ M ∩ N and π ∈ RM (N), the inductive hypothesis implies that

Aη,π ∩ κ = AαM,N ,π ∩ κ ⊆ π. So γ < π < ζ. This completes the induction. In particular,

Q0 ∩ κ = AαM,N ,σ ∩ κ ⊆ σ. 52 THOMAS GILTON AND JOHN KRUEGER

Conversely, let β < σ, and we will show that β ∈ Q0. Fix η ∈ lim(CαM,N )∩(M ∩N) 0 with κ ≤ η. Then by Lemma 9.6, there is β < σ such that β ∈ Aη,β0 . So

0 0 0 β ∈ Aη,β = AαM,N ,β ∩ η ⊆ AαM,N ,β ⊆ AαM,N ,σ ⊆ Q0. Lemma 9.8. Let M and N be in X , where {M,N} is adequate and N is simple. Suppose that σ ∈ RN (M), ζ ∈ RM (N), and ζ = min((N ∩ κ) \ σ). Let η := + min((N ∩ κ ) \ αM,N ). Let Q0 := Q0(M, N, σ) and Q := Q(N, M, ζ). Then: (1) Q ∈ N ∩ Y; + (2) Q ∩ κ = ζ, Q ∩ κ = Aη,ζ , and sup(Q) = η; + + (3) N ∩ Q ∩ κ = N ∩ Q0 ∩ κ ; (4) M ∩ N ∩ κ+ ⊆ Q.

Proof. We will apply Lemma 9.2 to the models N and Q0. Let us check that the assumptions of this lemma hold, using Lemma 9.7. We know that N ∈ X is simple, Q0 ∈ Y, and Q0 ∩ κ = σ < ζ < sup(N ∩ κ).

Also, sup(N ∩ Q0) = αM,N , since sup(Q0) = αM,N , sup(M ∩ N) = αM,N , and + M ∩ N ∩ κ ⊆ N ∩ Q0. Moreover,

min((N ∩ κ) \ (Q0 ∩ κ)) = min((N ∩ κ) \ σ) = ζ, and + + min((N ∩ κ ) \ sup(N ∩ Q0)) = min((N ∩ κ ) \ αM,N ) = η. By Notations 9.1 and 9.3,

Q(N,Q0) = Sk(Aη,ζ ) = Q(N, M, ζ) = Q. By Lemma 9.2, we have that: (a) Q ∈ N ∩ Y; + (b) Q ∩ κ = ζ, Q ∩ κ = Aη,ζ , and sup(Q) = η; + + (c) N ∩ Q ∩ κ = N ∩ Q0 ∩ κ . + + This proves (1), (2), and (3). By Lemma 9.7(3), M ∩N ∩κ ⊆ Q0. So M ∩N ∩κ ⊆ + + N ∩ Q0 ∩ κ = N ∩ Q ∩ κ ⊆ Q, which proves (4). The next lemma summarizes Lemmas 9.4 and 9.8. Lemma 9.9. Let M and N be in X such that {M,N} is adequate and N is simple. + Let ζ ∈ RM (N), η := min((N ∩ κ ) \ αM,N ), and Q := Q(N, M, ζ). Then: (1) Q ∈ N ∩ Y; + (2) Q ∩ κ = ζ, Q ∩ κ = Aη,ζ , and sup(Q) = η; (3) M ∩ N ∩ κ+ ⊆ Q; + + (4) if ζ = min((N ∩ κ) \ βM,N ), then N ∩ Q ∩ κ = M ∩ N ∩ κ ; + (5) if ζ = min((N ∩ κ) \ σ), where σ ∈ RN (M), then N ∩ Q ∩ κ = N ∩ + Q0(M, N, σ) ∩ κ . Proof. Immediate from Lemmas 9.4 and 9.8. Let us derive some additional information about the model Q(N, M, ζ). Lemma 9.10. Let M and N be in X such that {M,N} is adequate and N is + simple. Let ζ ∈ RM (N), η := min((N ∩ κ ) \ αM,N ), and Q := Q(N, M, ζ). Then:

(1) if P ∈ M ∩ Y and sup(N ∩ ζ) < P ∩ κ < ζ, then N ∩ P ∩ αM,N ⊆ Q; MITCHELL’S THEOREM REVISITED 53

(2) if N ≤ M, P ∈ M ∩ Y, and P ∩κ < sup(M ∩N ∩κ), then N ∩P ∩κ+ ⊆ Q; (3) if M < N and P ∈ M ∩ N ∩ Y, then N ∩ P ∩ κ+ ⊆ Q.

Proof. Note that since αM,N is a limit point of N, for all ξ < κ, AαM,N ,ξ = Aη,ξ ∩ αM,N by Lemma 7.12. (1) Suppose that P ∈ M ∩ Y and sup(N ∩ ζ) < P ∩ κ < ζ. Since βM,N ≤ ζ,

M ∩ N ∩ κ = M ∩ N ∩ βM,N ⊆ N ∩ ζ. So sup(M ∩ N ∩ κ) ≤ sup(N ∩ ζ) < P ∩ κ. By Lemma 8.10,

P ∩ N ∩ αM,N ⊆ AαM,N ,P ∩κ ⊆ AαM,N ,ζ = Aη,ζ ∩ αM,N ⊆ Q. (2) If N ≤ M, P ∈ M ∩ Y, and P ∩ κ < sup(M ∩ N ∩ κ), then N ∩ P ∩ κ+ ⊆ M by Lemma 8.7. Hence N ∩ P ∩ κ+ ⊆ M ∩ N ∩ κ+ ⊆ Q by Lemma 9.9(3). (3) Suppose that M < N and P ∈ M ∩ N ∩ Y. Then sup(P ) and P ∩ κ are in + M ∩ N ∩ κ by elementarity, and hence in Q by Lemma 9.9(3). So Asup(P ),P ∩κ = + + + P ∩ κ ∈ Q by elementarity. So P ∩ κ ⊆ Q. In particular, N ∩ P ∩ κ ⊆ Q.

+ Finally, we consider canonical models determined by ordinals in RN (M). Notation 9.11. Let M and N be in X , where {M,N} is adequate and N is simple. + Let ζ ∈ RM (N) and σ ∈ RN (M). Let X be any nonempty set of P ∈ M ∩ Y such that sup(N ∩ ζ) < P ∩ κ < ζ and P ∩ N ∩ [sup(M ∩ σ), σ) 6= ∅. We let PX denote S the set Sk( {P ∩ σ : P ∈ X}) and βX denote the ordinal PX ∩ κ. Lemma 9.12. Under the assumptions of Notation 9.11, the following statements hold:

(1) PX ∈ Y; (2) βX = sup{P ∩ κ : P ∈ X} < ζ; + S (3) PX ∩ κ = {P ∩ σ : P ∈ X}; (4) sup(PX ) = σ. + Proof. Let P ∈ X. Since σ ∈ RN (M), σ ∈ M and σ has uncountable coﬁnality. Also P ∈ M and P ∩ [sup(M ∩ σ), σ) 6= ∅, which imply that σ is a limit point of P by Lemma 7.30. It follows that if P1 and P2 are in X and P1 ∩ κ ≤ P2 ∩ κ, then P1 ∩ σ ⊆ P2 ∩ σ by Lemma 7.28. Thus {P ∩ σ : P ∈ X} is a subset increasing sequence. Since each P ∩ σ is closed under H∗ by Lemma 8.16, the set S{P ∩ σ : P ∈ X} is closed under H∗. Hence

+ [ + [ PX ∩ κ = Sk( {P ∩ σ : P ∈ X}) ∩ κ = {P ∩ σ : P ∈ X}, which proves (3). Since σ is a limit point of P ∩ σ for each P ∈ X, obviously σ is a limit point of + PX . But PX ∩ κ ⊆ σ, so sup(PX ) = σ, which proves (4). Clearly

βX = PX ∩ κ = sup{P ∩ κ : P ∈ X}, which is in κ. Since P ∩ κ < ζ for all P ∈ X, it follows that PX ∩ κ ≤ ζ. But P ∩ κ ∈ M ∩ κ for all P ∈ X. Therefore PX ∩ κ ∈ cl(M ∩ κ), which implies that βX = PX ∩ κ < ζ by Lemma 2.2(1), which proves (2). 54 THOMAS GILTON AND JOHN KRUEGER

To show that PX ∈ Y, it suffices to show that lim(Cσ) ∩ PX is cofinal in σ. Fix P ∈ X. Then it will suffice to show that lim(Cσ) ∩ P is cofinal in σ, since this set is a subset of PX . First, assume that σ∈ / P . Then σ ∈ cl(P ) \ P , which implies by Lemma 7.13 that lim(Cσ)∩P is cofinal in σ. Secondly, assume that σ ∈ P . Since σ is a limit point of P and |P | < κ, cf(σ) < κ. So ot(Cσ) < κ. Hence ot(Cσ) ∈ P ∩ κ and P ∩ κ ∈ κ, which implies that Cσ ⊆ P . As σ has uncountable cofinality, clearly lim(Cσ) is cofinal in σ. So lim(Cσ) ∩ P is cofinal in σ. Notation 9.13. Under the assumptions of Notation 9.11, we let

Q(N, M, ζ, σ, X) := Sk(AηN ,ζ ), where η := sup(N ∩ PX ).

Note that PX ∈ Y and βX = PX ∩ κ < ζ < sup(N ∩ κ) imply by Lemma 8.13 that ηN exists. Also since ζ = min((N ∩ κ) \ βX ), Q(N, M, ζ, σ, X) is equal to Q(N,PX ) from Notation 9.1. Lemma 9.14. Let M and N be in X , where {M,N} is adequate and N is simple. + Let ζ ∈ RM (N) and σ ∈ RN (M). Let X be any nonempty set of P ∈ M ∩ Y such that sup(N ∩ ζ) < P ∩ κ < ζ and P ∩ N ∩ [sup(M ∩ σ), σ) 6= ∅. Let η := sup(N ∩ PX ) and Q := Q(N, M, ζ, σ, X). Then: (1) Q ∈ N ∩ Y; + (2) Q ∩ κ = ζ, Q ∩ κ = AηN ,ζ , and sup(Q) = ηN ; + + (3) N ∩ Q ∩ κ = N ∩ PX ∩ κ ; (4) for all P ∈ X, N ∩ P ∩ σ ⊆ Q.

Proof. As noted above, Q = Q(N,PX ). Also η = sup(N ∩ PX ) and ζ = min((N ∩ κ) \ (PX ∩ κ)). By Lemma 9.2: (a) Q ∈ N ∩ Y; + (b) Q ∩ κ = ζ, Q ∩ κ = AηN ,ζ , and sup(Q) = ηN ; + + (c) N ∩ PX ∩ κ = N ∩ Q ∩ κ . This proves (1), (2), and (3). In particular, if P ∈ X, then + N ∩ P ∩ σ ⊆ N ∩ PX ∩ κ ⊆ Q, which proves (4).

§10. Closure under canonical models

+ Fix a sequence hSη : η < κ i, where each Sη is a subset of κ ∩ cof(> ω). Let us assume that the structure A from Notation 7.6 includes S~ as a predicate. In this section we will show that we can add canonical models to an S~-obedient side condition and preserve S~-obediency. As stated in the comments prior to Deﬁnition 5.2, the deﬁnitions of S~-adequate and S~-obedient are made relative to a subclass of Y0. For the remainder of the paper, this subclass will be the set Y from Notation 7.8. Lemma 10.1. Let (A, B) be an S~-obedient side condition. Suppose that N ∈ A is simple. Let P ∈ B be such that P ∩ κ < sup(N ∩ κ). Let Q := Q(N,P ). Then (A, B ∪ {Q}) is an S~-obedient side condition. MITCHELL’S THEOREM REVISITED 55

See Notation 9.1 for the deﬁnition of Q(N,P ). Proof. Let β := P ∩ κ. By Lemma 9.2, + + Q ∈ N ∩ Y,Q ∩ κ = βN , and N ∩ Q ∩ κ = N ∩ P ∩ κ . Let us show that Q is S~-strong. Since Q ∈ N, it suﬃces to show that N models + that Q is S~-strong. Let τ ∈ N ∩ Q ∩ κ , and we will show that Q ∩ κ = βN ∈ Sτ . But τ ∈ N ∩ Q ∩ κ+ = N ∩ P ∩ κ+. Since N ∈ A, P ∈ B, and (A, B) is S~-obedient, it follows that βN ∈ Sτ . + Let M ∈ A, and suppose that ζ = min((M ∩ κ) \ βN ). Fix τ ∈ M ∩ Q ∩ κ , and we will show that ζ ∈ Sτ . If ζ = βN , then ζ ∈ Sτ because Q is S~-strong. Assume that βN < ζ, which means that βN ∈/ M. First assume that ζ ∈ RN (M). Then since Q ∈ N ∩ Y is S~-strong and sup(M ∩ ζ) < Q ∩ κ = βN < ζ, it follows that ζ ∈ Sτ as A is S~-adequate. In particular, if βM,N ≤ βN , then ζ ∈ RN (M). Suppose that βN < βM,N ≤ ζ. Then ζ = min((M ∩ κ) \ βM,N ). Since βN ∈ (N ∩ βM,N ) \ M, we have that M < N. So ζ = min((M ∩ κ) \ βM,N ) is in RN (M). The remaining case is that ζ < βM,N . Then since βN ∈ (N ∩ βM,N ) \ M, it follows that M < N. So

M ∩ ζ ⊆ M ∩ βM,N ⊆ N. As τ ∈ M ∩ Q ∩ κ+, Q ∈ N ∩ Y, and

Q ∩ κ < ζ < sup(M ∩ βM,N ) = sup(M ∩ N ∩ κ), it follows that τ ∈ N by Lemma 8.7. So τ ∈ N ∩ Q ∩ κ+ = N ∩ P ∩ κ+. Since M ∩ ζ ⊆ N and ζ = min((M ∩ κ) \ βN ), we have that ζ = min((M ∩ κ) \ β) = min((M ∩ κ) \ (P ∩ κ)). + Since M ∈ A, P ∈ B, and τ ∈ M ∩ P ∩ κ , it follows that ζ ∈ Sτ as (A, B) is ~ S-obedient. Lemma 10.2. Let (A, B) be an S~-obedient side condition. Let N ∈ A be simple and M ∈ A. Suppose that N ≤ M and ζ = min((N ∩ κ) \ βM,N ). Let Q := Q(N, M, ζ). Then (A, B ∪ {Q}) is an S~-obedient side condition. See Notation 9.3 for the definition of Q(N, M, ζ). Proof. By Lemma 9.4, Q ∈ N ∩ Y,Q ∩ κ = ζ, and N ∩ Q ∩ κ+ = M ∩ N ∩ κ+. First we show that Q is S~-strong. Since Q ∈ N, it suffices to show that N models + that Q is S~-strong. Let τ ∈ N ∩ Q ∩ κ , and we will show that Q ∩ κ = ζ is in Sτ . + + Then τ ∈ N ∩ Q ∩ κ = M ∩ N ∩ κ . So τ ∈ M ∩ N. Since ζ ∈ RM (N), ζ ∈ Sτ as A is S~-adequate. Now let K ∈ A, and suppose that θ = min((K ∩ κ) \ ζ). Fix τ ∈ K ∩ Q ∩ κ+, and we will show that θ ∈ Sτ . If ζ = θ, then θ ∈ Sτ since Q is S~-strong. So assume that ζ < θ, which means that ζ∈ / K. Suppose first that θ ∈ RN (K). Then since Q ∈ N ∩ Y is S~-strong and sup(K ∩ θ) < Q ∩ κ = ζ < θ, it follows that θ ∈ Sτ as A is S~-adequate. In particular, if βK,N ≤ ζ, then θ ∈ RN (K). Suppose that ζ < βK,N ≤ θ. Then θ = min((K ∩ κ) \ 56 THOMAS GILTON AND JOHN KRUEGER

βK,N ). Since ζ ∈ (N ∩βK,N )\K, we have that K < N. So θ = min((K ∩κ)\βK,N ) is in RN (K). The remaining case is that θ < βK,N . We apply Lemma 2.7. We have that {K,M,N} is adequate, ζ ∈ RM (N), ζ∈ / K, θ = min((K ∩ κ) \ ζ), and θ < βK,N . By Lemma 2.7, θ ∈ RM (K). Since ζ ∈ (N ∩ βK,N ) \ K, it follows that K < N. As Q ∈ N ∩ Y, K < N, τ ∈ K ∩ Q, and

Q ∩ κ = ζ < θ < sup(K ∩ βK,N ) = sup(K ∩ N ∩ κ), it follows that τ ∈ N by Lemma 8.7. So τ ∈ N ∩ Q ∩ κ+ = M ∩ N ∩ κ+. Hence ~ τ ∈ K ∩ M. Since θ ∈ RM (K), it follows that θ ∈ Sτ as A is S-adequate. Lemma 10.3. Let M and N be in X such that {M,N} is adequate and N is simple. Assume that M ≺ (A, Y). Let σ ∈ RN (M) and ζ ∈ RM (N). Then Q0(M, N, σ) and Q(N, M, ζ) are S~-strong. Recall that (A, Y) is the structure A augmented with the additional predicate Y. See Notations 9.3 and 9.5 for the deﬁnitions of Q(N, M, ζ) and Q0(M, N, σ).

Proof. The proof is by induction on remainder points in RM (N) ∪ RN (M). First consider ζ ∈ RM (N). If ζ = min((N ∩ κ) \ βM,N ), then Q(N, M, ζ) is S~-strong by Lemma 10.2. So assume that ζ = min((N ∩ κ) \ σ), for some σ ∈ RN (M). Let Q := Q(N, M, ζ) and Q0 := Q0(M, N, σ). By the inductive hypothesis, Q0 is S~-strong. And by Lemma 9.7, + Q0 ∩ κ = σ and Q0 ∩ κ = AαM,N ,σ. To show that Q is S~-strong, it suﬃces to prove that N models that Q is S~-strong. + Let τ ∈ N ∩ Q ∩ κ , and we will show that Q ∩ κ ∈ Sτ . By Lemma 9.8, + Q ∩ κ = ζ and Q ∩ κ = Aη,ζ , + where η := min((N ∩ κ ) \ αM,N ), and + + N ∩ Q ∩ κ = N ∩ Q0 ∩ κ .

In particular, τ ∈ N ∩ Q0. Also

τ ∈ N ∩ Aη,ζ ⊆ αM,N , so τ < αM,N .

Fix θ ∈ lim(CαM,N ) ∩ (M ∩ N) greater then τ. Then

τ ∈ Q0 ∩ θ = AαM,N ,σ ∩ θ = Aθ,σ.

Since θ and σ are in M, Q0 is S~-strong, Q0 ∩ κ = σ, and Aθ,σ ⊆ Q0, by the elementarity of M we can ﬁx an S~-strong model P ∈ M ∩ Y such that P ∩ κ = σ and Aθ,σ ⊆ P . Then τ ∈ N ∩P . Since ζ ∈ RM (N) and sup(N ∩ζ) < σ = P ∩κ < ζ, it follows that ζ ∈ Sτ as A is S~-adequate. Now consider σ ∈ RN (M), and we will show that Q0 := Q0(M, N, σ) is S~-strong.

We ﬁrst claim that for all θ ∈ lim(CαM,N ) ∩ (M ∩ N) and for all τ ∈ Aθ,σ, σ ∈ Sτ .

So fix θ ∈ lim(CαM,N ) ∩ (M ∩ N). Since θ and σ are in M, it suffices to prove that M models that for all τ ∈ Aθ,σ, σ ∈ Sτ . Let τ ∈ M ∩ Aθ,σ. Since σ is a limit ordinal, by elementarity we can fix γ ∈ M ∩ σ such that τ ∈ Aθ,γ . MITCHELL’S THEOREM REVISITED 57

If σ = min((M ∩ κ) \ βM,N ), then M ≤ N and γ ∈ M ∩ βM,N ⊆ N. So θ is in N, γ < sup(M ∩ N ∩ κ), and τ ∈ M ∩ Aθ,γ , which by Lemma 8.6 implies that τ ∈ N. + So τ ∈ M ∩ N ∩ κ . As σ ∈ RN (M), it follows that σ ∈ Sτ as A is S~-adequate. Otherwise there is ζ ∈ RM (N) such that σ = min((M ∩ κ) \ ζ). By the inductive hypothesis, Q := Q(N, M, ζ) is S~-strong. Since αM,N is a limit point of M ∩ N and + M ∩ N ∩ κ ⊆ Q by Lemma 9.9(3), it follows that αM,N is a limit point of Q. So

Q ∩ αM,N = AαM,N ,Q∩κ by Lemma 7.27. By Lemma 9.9(2), Q ∩ κ = ζ. So

Q ∩ αM,N = AαM,N ,ζ . Now γ ∈ M ∩ σ ⊆ ζ. Hence

τ ∈ M ∩ Aθ,γ ⊆ M ∩ Aθ,ζ = M ∩ AαM,N ,ζ ∩ θ ⊆ M ∩ Q. So we have that Q ∈ N ∩Y is S~-strong, sup(M ∩σ) < ζ = Q∩κ < σ, and τ ∈ M ∩Q.

Since σ ∈ RN (M), it follows that σ ∈ Sτ as A is S~-adequate.

This completes the proof of the claim that for all θ ∈ lim(CαM,N ) ∩ (M ∩ N), for + all τ ∈ Aθ,σ, σ ∈ Sτ . Now we show that Q0 is S~-strong. By Lemma 9.7, Q0 ∩ κ = + AαM,N ,σ. Let τ ∈ Q0 ∩ κ . Then τ < αM,N . Fix θ ∈ lim(CαM,N ) ∩ (M ∩ N) which is greater than τ. Then

τ ∈ Q0 ∩ θ = AαM,N ,σ ∩ θ = Aθ,σ.

By the claim, σ ∈ Sτ . Lemma 10.4. Let (A, B) be an S~-obedient side condition. Suppose that N ∈ A is simple, M ∈ A, and M ≺ (A, Y). Let ζ ∈ RM (N). Let Q := Q(N, M, ζ). Then (A, B ∪ {Q}) is an S~-obedient side condition.

Proof. If ζ = min((N ∩ κ) \ βM,N ), then we are done by Lemma 10.2. So assume that ζ = min((N ∩ κ) \ σ), where σ ∈ RN (M). Let Q0 := Q0(M, N, σ). By Lemma 9.8, + + Q ∩ κ = ζ and N ∩ Q ∩ κ = N ∩ Q0 ∩ κ . By Lemma 9.7, + Q0 ∩ κ = σ and Q0 ∩ κ = AαM,N ,σ.

Also Q0 and Q are S~-strong by Lemma 10.3. Suppose that K ∈ A and θ = min((K ∩ κ) \ ζ). Let τ ∈ K ∩ Q ∩ κ+, and we will show that θ ∈ Sτ . If ζ = θ, then θ ∈ Sτ since Q is S~-strong. So assume that ζ < θ, which means that ζ∈ / K. First consider the case that θ ∈ RN (K). Then since sup(K ∩ θ) < ζ = Q ∩ κ < θ and Q ∈ N ∩Y is S~-strong, it follows that θ ∈ Sτ as A is S~-adequate. In particular, if βK,N ≤ ζ, then θ ∈ RN (K). Suppose that ζ < βK,N ≤ θ. Then θ = min((K ∩κ)\ βK,N ). Since ζ ∈ (N ∩βK,N )\K, we have that K < N. So θ = min((K ∩κ)\βK,N ) is in RN (K). The remaining case is that θ < βK,N . We apply Lemma 2.7. We have that {K,M,N} is adequate, ζ ∈ RM (N), ζ∈ / K, θ = min((K ∩ κ) \ ζ), and θ < βK,N . 58 THOMAS GILTON AND JOHN KRUEGER

By Lemma 2.7, θ ∈ RM (K). Since ζ ∈ (N ∩ βK,N ) \ K, we have that K < N. As Q ∈ N ∩ Y,

Q ∩ κ = ζ < θ < sup(K ∩ βK,N ) = sup(K ∩ N ∩ κ), and τ ∈ K ∩ Q, it follows that τ ∈ N by Lemma 8.7. So + + τ ∈ N ∩ Q ∩ κ = N ∩ Q0 ∩ κ . + Hence τ ∈ K ∩ Q0. Since Q0 ∩ κ = AαM,N ,σ, it follows that τ < αM,N .

Fix π ∈ lim(CαM,N ) ∩ (M ∩ N) with τ < π. Then

τ ∈ Q0 ∩ π = AαM,N ,σ ∩ π = Aπ,σ.

Since π and σ are in M, Q0 is S~-strong, Q0 ∩ κ = σ, and Aπ,σ ⊆ Q0, by the elementarity of M we can ﬁx P ∈ M ∩ Y which is S~-strong such that P ∩ κ = σ and Aπ,σ ⊆ P . In particular, τ ∈ P . Since K ∩ θ ⊆ N and ζ = min((N ∩ κ) \ σ), clearly θ = min((K ∩ κ) \ σ). So τ ∈ K ∩ P , P ∈ M ∩ Y is S~-strong, and sup(K ∩ θ) < σ = P ∩ κ < θ. Since θ ∈ RM (K), it follows that θ ∈ Sτ as A is ~ S-adequate. Notation 10.5. Let M and N be in X , where {M,N} is adequate and N is simple. + Let ζ ∈ RM (N) and σ ∈ RN (M). Let X be the set of P ∈ M ∩ Y such that P is S~-strong, sup(N ∩ ζ) < P ∩ κ < ζ, and P ∩ N ∩ [sup(M ∩ σ), σ) 6= ∅. Assume that X is nonempty. We let Q(N, M, ζ, σ, S~) denote the set Q(N, M, ζ, σ, X). See Notation 9.13 for the deﬁnition of Q(N, M, ζ, σ, X). Lemma 10.6. Let (A, B) be an S~-obedient side condition. Let M and N be in A, + ~ where N is simple. Let ζ ∈ RM (N) and σ ∈ RN (M). Let Q := Q(N, M, ζ, σ, S). Then (A, B ∪ {Q}) is an S~-obedient side condition.

Proof. Let X be as in Notation 10.5, and let PX be as in Notation 9.11. Then by Lemma 9.14, + + Q ∈ N ∩ Y,Q ∩ κ = ζ, and N ∩ Q ∩ κ = N ∩ PX ∩ κ . Let us prove that Q is S~-strong. Since Q ∈ N, it suﬃces to show that N models + that Q is S~-strong. Fix τ ∈ N ∩ Q ∩ κ , and we will show that Q ∩ κ = ζ ∈ Sτ . + + Since N ∩ Q ∩ κ = N ∩ PX ∩ κ , we have that τ ∈ PX . By the deﬁnition of PX , for some P ∈ X, τ ∈ P . But then sup(N ∩ ζ) < P ∩ κ < ζ, P ∈ M ∩ Y is S~-strong, and τ ∈ N ∩ P . Since ζ ∈ RM (N), this implies that ζ ∈ Sτ as A is S~-adequate. + Let K ∈ Ap, and suppose that θ = min((K ∩ κ) \ ζ). Fix τ ∈ K ∩ Q ∩ κ , and we will show that θ ∈ Sτ . If θ = ζ, then θ ∈ Sτ since Q is S~-strong. So assume that ζ < θ, which means that ζ∈ / K. If θ ∈ RN (K), then since Q ∈ N ∩ Y is S~-strong, sup(K ∩ θ) < Q ∩ κ < θ, and τ ∈ K ∩ Q, it follows that θ ∈ Sτ as A is S~-adequate. In particular, if βK,N ≤ ζ, then θ ∈ RN (K). Suppose that ζ < βK,N ≤ θ. Then θ = min((K ∩ κ) \ βK,N ). Since ζ ∈ (N ∩ βK,N ) \ K, we have that K < N, which implies that θ ∈ RN (K). The remaining case is that θ < βK,N . We apply Lemma 2.7. We have that {K,M,N} is adequate, ζ ∈ RM (N), ζ∈ / K, θ = min((K ∩ κ) \ ζ), and θ < βK,N . By Lemma 2.7, θ ∈ RM (K). Since ζ ∈ (N ∩ βK,N ) \ K, we have that K < N. As Q ∈ N ∩ Y,

Q ∩ κ = ζ < θ < sup(K ∩ βK,N ) = sup(K ∩ N ∩ κ), MITCHELL’S THEOREM REVISITED 59 and τ ∈ K ∩ Q, it follows that τ ∈ N by Lemma 8.7. So

+ + τ ∈ N ∩ Q ∩ κ = N ∩ PX ∩ κ .

By the deﬁnition of PX , there is P ∈ X such that τ ∈ P . Since sup(N ∩ζ) < P ∩κ < ζ and K ∩ θ ⊆ N, clearly sup(K ∩ θ) < P ∩ κ < θ. As P ∈ M ∩ Y is S~-strong, ~ τ ∈ K ∩ P , and θ ∈ RM (K), it follows that θ ∈ Sτ since A is S-adequate.

Deﬁnition 10.7. Let (A, B) be an S~-obedient side condition. Suppose that N ∈ A is simple. We say that (A, B) is closed under canonical models with respect to N if: (1) for all P ∈ B with P ∩ κ < sup(N ∩ κ), Q(N,P ) ∈ B; (2) for all M ∈ A and ζ ∈ RM (N), Q(N, M, ζ) ∈ B; + ~ (3) for all M ∈ A, ζ ∈ RM (N), and σ ∈ RN (M), Q(N, M, ζ, σ, S) ∈ B. Proposition 10.8. Let (A, B) be an S~-obedient side condition such that for all M ∈ A, M ≺ (A, Y). Suppose that N ∈ A is simple. Then there exists (A, C) such that B ⊆ C, (A, C) is an S~-obedient side condition, and (A, C) is closed under canonical models with respect to N.

Proof. First apply Lemma 10.1 finitely many times to obtain C0 such that B ⊆ C0,(A, C0) is an S~-obedient side condition, and (A, C0) satisfies property (1) of Definition 10.7. Then apply Lemmas 10.4 and 10.6 finitely many times to obtain C such that C0 ⊆ C,(A, C) is an S~-obedient side condition, and (A, C) satisfies properties (2) and (3) of Definition 10.7. Since all of the models which are added are in N, and for all P ∈ N ∩ Y, Q(N,P ) = P , it follows that (A, C) also satisfies property (1) of Definition 10.7.

Lemma 10.9. Suppose that (A, B) is an S~-obedient side condition, and N ∈ A is simple. Assume that (A, B) is closed under canonical models with respect to N. Then: (1) Suppose that P ∈ B, P ∩ κ < sup(N ∩ κ), and τ ∈ N ∩ P ∩ κ+. Then there is Q ∈ B ∩ N such that Q ∩ κ = min((N ∩ κ) \ (P ∩ κ)) and τ ∈ Q. (2) Suppose that M ∈ A and ζ ∈ RM (N). Then there is Q ∈ B ∩ N such that Q ∩ κ = ζ and M ∩ N ∩ κ+ ⊆ Q. (3) Suppose that M ∈ A, M < N, and ζ ∈ RM (N). Then there is Q ∈ B ∩ N such that Q ∩ κ = ζ, and for all P ∈ M ∩ N ∩ Y which is S~-strong, N ∩ P ∩ κ+ ⊆ Q. (4) Suppose that M ∈ A, ζ ∈ RM (N), P ∈ M ∩ Y is S~-strong, sup(N ∩ ζ) < P ∩ κ < ζ, and τ ∈ N ∩ P ∩ κ+. Then there is Q ∈ B ∩ N such that Q ∩ κ = ζ and τ ∈ Q. Proof. (1) Suppose that P ∈ B,P ∩ κ < sup(N ∩ κ), and τ ∈ N ∩ P ∩ κ+. Then Q(N,P ) ∈ B ∩ N. By Lemma 9.2, Q(N,P ) ∩ κ = min((N ∩ κ) \ (P ∩ κ)) and N ∩ P ∩ κ+ ⊆ Q(N,P ). In particular, τ ∈ Q(N,P ). 60 THOMAS GILTON AND JOHN KRUEGER

(2,3) Let M ∈ A and ζ ∈ RM (N). Let Q := Q(N, M, ζ). Then Q ∈ B ∩ N. By Lemma 9.9, Q ∩ κ = ζ and M ∩ N ∩ κ+ ⊆ Q, which proves (2). If in addition M < N, then by Lemma 9.10(3), for all P ∈ M ∩ N ∩ Y, N ∩ P ∩ κ+ ⊆ Q, which proves (3).

(4) Suppose that M ∈ A, ζ ∈ RM (N), P ∈ M ∩ Y is S~-strong, sup(N ∩ ζ) < P ∩ + κ < ζ, and τ ∈ N ∩P ∩κ . First assume that τ < αM,N . Then Q(N, M, ζ) ∈ B∩N, and Q(N, M, ζ) ∩ κ = ζ by Lemma 9.9. Also by Lemma 9.10(1),

N ∩ P ∩ αM,N ⊆ Q(N, M, ζ). Hence τ ∈ Q(N, M, ζ). Assume that αM,N ≤ τ. Note that σ := τM exists since τ < sup(P ) ∈ M. As + ~ τ ∈ N, σ is in RN (M). So τ ∈ N ∩ P ∩ σ. Let Q := Q(N, M, ζ, σ, S), which is in B ∩ N. Then Q ∩ κ = ζ and N ∩ P ∩ σ ⊆ Q by Lemma 9.14. In particular, τ ∈ Q.

§11. The main proxy lemma

Let M ∈ X and N ∈ X ∪ Y, where N is simple. Suppose that M < N in the case that N ∈ X , and sup(M ∩ N ∩ κ) < N ∩ κ in the case that N ∈ Y. Consider P ∈ M ∩ Y such that P ∩ κ < sup(M ∩ N ∩ κ), and assume that we are building an object in N which needs to be compatible in some sense with the model P . By Lemma 8.2, we know that M ∩ N is a member of N. However, when we intersect M with N, the model P will disappear if it is not in N. Thus although N sees a fragment of M, it does not necessarily see P even though P ∩ κ is in N. Proxies are designed to handle this situation. We will deﬁne an object p(M,N), called the canonical proxy of M and N, which is a member of N. The canonical proxy codes enough information about M that we can rebuild fragments of P inside N which can be used to avoid incompatibilities between P and the object we are constructing.7 Although the description and the proof of the existence of proxies is quite complicated, in practice when we use proxies we only need to appeal to a single result, called the main proxy lemma, which is Lemma 11.5 below. In applications of proxies, it is not necessary to understand anything else about proxies except what is contained in that lemma. The next lemma asserts the existence of proxies. We will postpone the proof until the next section.

7The idea of a canonical proxy which we use in this paper is a variation of a technical device used by Mitchell for a similar purpose. In the proof of Mitchell’s theorem from [12], a side condition is a pair (M, a), where M is a countable model and a is a proxy. In this paper we separate the idea of a side condition and a proxy. In contrast to [12], where proxies are present in many diﬀerent parts of the proof, all applications of proxies which we give reduce to a single lemma, which is the main proxy lemma, Lemma 11.5. The idea of a canonical proxy and the main proxy lemma are new to this paper and do not appear in [12]. MITCHELL’S THEOREM REVISITED 61

Lemma 11.1 (Proxy existence lemma). Let M ∈ X and N ∈ X ∪ Y, where N is simple. Assume that M < N in the case that N ∈ X , and sup(M ∩ N ∩ κ) < N ∩ κ ∗ + 0 in the case that N ∈ Y. Let η ∈ RN (M). Then there exist finite sets a and a satisfying the following statements: (1) a is a finite set of pairs of ordinals, and a0 = {σ : ∃β (β, σ) ∈ a}. (2) For all (β, σ) in a, (a) β ∈ M ∩ N ∩ κ; (b) σ ∈ N ∩ κ+ is a limit ordinal; (c) sup(N ∩ σ) ≤ η∗; (d) if a 6= ∅, then min(a0) = min((N ∩ κ+) \ sup(M ∩ η∗)). (3) If (β, σ) ∈ a, where min(a0) < σ, and β ≤ γ < κ, then: (a) Aη∗,γ ∩ sup(N ∩ σ) = Aσ,γ ∩ sup(N ∩ σ); (b) Aη∗,γ ∩ N ∩ σ = Aσ,γ ∩ N. (4) If P ∈ M ∩ Y, P ∩ κ ∈ M ∩ N ∩ κ, and P ∩ N ∩ [sup(M ∩ η∗), η∗) 6= ∅, then there exists σ ∈ a0 such that: (a) P ∩ N ∩ η∗ ⊆ σ; (b) the least such σ is equal to the largest σ in a0 such that for some β, β ≤ P ∩ κ and (β, σ) ∈ a. (5) Let P and σ be as in (4), and assume that (β, σ) ∈ a; then: (a) β ≤ P ∩ κ; (b) P ∩ sup(N ∩ σ) = Aσ,P ∩κ ∩ sup(N ∩ σ); ∗ (c) P ∩ N ∩ η = Aσ,P ∩κ ∩ N. For the remainder of this section, we will assume that the proxy existence lemma holds. We now define the canonical proxy p(M,N). A lexicographical ordering on sets of pairs of ordinals is described as follows. We identify a finite set of pairs of ordinals as a finite set of ordinals using the Gödelpairing function, and then compare any two finite sets of pairs using the lexicographical ordering on their corresponding sets of ordinals. Definition 11.2. Let M ∈ X and N ∈ X ∪ Y, where N is simple. Assume that M < N in the case that N ∈ X , and sup(M ∩ N ∩ κ) < N ∩ κ in the case that N ∈ Y. + Let η0, . . . , ηk−1 enumerate the ordinals in RN (M) in increasing order. Define p(M,N) as the function with domain k such that for all i < k, p(M,N)(i) is the ∗ lexicographically least set a satisfying (1)–(5) of Lemma 11.1 for η = ηi. Note that p(M,N) is a member of N. The proof of the main proxy lemma will use the next two technical lemmas. Lemma 11.3. Let M ∈ X and N ∈ X ∪ Y, where N is simple. Assume that M < N in the case that N ∈ X , and sup(M ∩ N ∩ κ) < N ∩ κ in the case that N ∈ Y. + ∗ + Let k be the size of RN (M), and assume that η is the i-th member of RN (M), where i < k. Let a := p(M,N)(i) and a0 := {σ : ∃β (β, σ) ∈ a}. Suppose that P ∈ M ∩ Y, P ∩ κ ∈ M ∩ N ∩ κ, cf(P ∩ κ) > ω, and P ∩ N ∩ [sup(M ∩ η∗), η∗) 6= ∅. Let σ be the least ordinal in a0 such that P ∩ N ∩ η∗ ⊆ σ, which exists by Lemma 11.1(4). Let Q := Sk(Aσ,P ∩κ). Then: (1) Q ∈ N ∩ Y; + (2) Q ∩ κ = P ∩ κ and Q ∩ κ = Aσ,P ∩κ; 62 THOMAS GILTON AND JOHN KRUEGER

(3) Q ∩ N ∩ κ+ = P ∩ N ∩ η∗; (4) Q ∩ sup(N ∩ σ) = P ∩ sup(N ∩ σ). In particular, P ∩ N ∩ η∗ ⊆ Q. Proof. Let θ := sup(N ∩ σ). By Lemma 11.1(5(b,c)), ∗ P ∩ θ = Aσ,P ∩κ ∩ θ and P ∩ N ∩ η = Aσ,P ∩κ ∩ N.

In particular, as Aσ,P ∩κ ∈ N, Lemmas 7.10 and 8.16 and the second equality imply ∗ + that Aσ,P ∩κ is closed under H . So Q ∩ κ = Aσ,P ∩κ. Hence + ∗ Q ∩ N ∩ κ = Aσ,P ∩κ ∩ N = P ∩ N ∩ η , which proves (3). Also by the ﬁrst equality,

Q ∩ sup(N ∩ σ) = Q ∩ θ = Aσ,P ∩κ ∩ θ = P ∩ θ = P ∩ sup(N ∩ σ), which proves (4). We claim that Q ∩ κ = P ∩ κ, which proves (2). Since Q and P ∩ κ are in N, it suffices to show that N models that Q∩κ = P ∩κ. So let α ∈ Q∩N ∩κ, and we will show that α < P ∩κ. Then α ∈ Q∩N ∩κ+ = P ∩N ∩η∗. So α ∈ P ∩κ. Conversely, let α ∈ N ∩P ∩κ, and we will show that α ∈ Q. Then α ∈ P ∩N ∩η∗ = Q∩N ∩κ+, so α ∈ Q. To prove (1), it suffices to show that lim(Csup(Q)) ∩ Q is cofinal in sup(Q). Since + ∗ ∗ Q ∩ κ = Aσ,P ∩κ, sup(Q) ≤ σ. Also note that since P ∩ [sup(M ∩ η ), η ) 6= ∅ and P and η∗ are in M, η∗ is a limit point of P by Lemma 7.30.

Case 1: θ < σ. Since cf(Q ∩ κ) = cf(P ∩ κ) > ω, it suffices by Lemma 7.15 to show that cf(sup(Q)) > ω. Since σ = min((N ∩ κ+) \ θ), σ has uncountable cofinality. So if sup(Q) = σ, then we are done. Otherwise by elementarity, sup(Q) ∈ N ∩σ ⊆ θ. By (4), Q∩κ+ = Q∩θ = P ∩θ. It follows that sup(Q) = sup(P ∩θ), which is a limit point of P below θ. Since η∗ is a limit point of P and sup(Q) < θ ≤ η∗ by Lemma 11.1(2(c)), if sup(Q) has countable cofinality then sup(Q) ∈ P by Lemma 7.14. But then sup(Q) ∈ P ∩ θ = Q ∩ θ, which is impossible. Therefore sup(Q) has uncountable cofinality.

Case 2: θ = σ. Then σ is a limit point of N, and in particular, σ has cofinality ω. By Lemma 11.1(2(c)), sup(N ∩ σ) = σ ≤ η∗. Since σ has cofinality ω and η∗ has uncountable cofinality, it follows that σ < η∗. We claim that sup(P ∩ N ∩ σ) < σ. Suppose for a contradiction that sup(P ∩ N ∩ σ) = σ. Then σ is a limit point of P . As η∗ is a limit point of P , σ < sup(P ). Since σ has cofinality ω and cf(P ∩ κ) > ω, Lemma 7.14 implies that σ ∈ P . So σ ∈ N ∩ P ∩ η∗, which contradicts that N ∩ P ∩ η∗ ⊆ σ. To show that Q ∈ Y, by Lemma 7.15 it suffices to show that cf(sup(Q)) > ω. Since θ = σ, by (4) we have that Q ∩ σ = P ∩ σ. Therefore Q ∩ N ∩ σ = P ∩ N ∩ σ. By the claim, sup(Q ∩ N ∩ σ) = sup(P ∩ N ∩ σ) < σ. If sup(Q) = σ, then since Q ∈ N and σ is a limit point of N, it is easy to argue by elementarity that Q ∩ N ∩ κ+ is cofinal in σ, which is false. Therefore sup(Q) < σ = θ. Since Q∩θ = P ∩θ, it follows that sup(Q) = sup(Q∩θ) = sup(P ∩θ), which is a limit point of P below θ. Since η∗ is a limit point of P and sup(Q) < θ ≤ η∗ by Lemma 11.1(2(c)), if sup(Q) has countable cofinality then sup(Q) ∈ P by Lemma MITCHELL’S THEOREM REVISITED 63

7.14. But then sup(Q) ∈ P ∩ θ = Q ∩ θ, which is impossible. Therefore sup(Q) has uncountable coﬁnality. Lemma 11.4. Let M ∈ X and N ∈ X ∪ Y, where N is simple. Assume that M < N in the case that N ∈ X , and sup(M ∩ N ∩ κ) < N ∩ κ in the case that N ∈ Y. + ∗ + Let k be the size of RN (M), and assume that η is the i-th member of RN (M), where i < k. Let a := p(M,N)(i) and a0 := {σ : ∃β (β, σ) ∈ a}. Suppose that (β, σ) ∈ a, where min(a0) < σ. Assume that Q ∈ N ∩ Y is such + that β ≤ Q ∩ κ, Q ∩ κ ∈ M ∩ N ∩ κ, cf(Q ∩ κ) > ω, Q ∩ κ = Aσ,Q∩κ, and Q ∩ N ∩ [sup(M ∩ η∗), σ) 6= ∅. Let P := Sk(Aη∗,Q∩κ). Then: (1) P ∈ M ∩ Y; + ∗ (2) P ∩ κ = Q ∩ κ, P ∩ κ = Aη∗,Q∩κ, and sup(P ) = η ; (3) P ∩ N ∩ [sup(M ∩ η∗), η∗) 6= ∅; (4) P ∩ M ∩ κ+ = Q ∩ M ∩ κ+. Proof. Let γ := Q ∩ κ and θ := sup(N ∩ σ). By Lemma 11.1(3),

Aη∗,γ ∩ θ = Aσ,γ ∩ θ and Aη∗,γ ∩ N ∩ σ = Aσ,γ ∩ N. We claim that Aη∗,γ ∩ M = Aσ,γ ∩ M. ∗ Let α ∈ Aη∗,γ ∩ M, and we will show that α ∈ Aσ,γ . Then α ∈ M ∩ η . By Lemma 11.1(2(d)), sup(M ∩ η∗) ≤ min(a0) < σ. Since min(a0) ∈ N, sup(M ∩ η∗) < sup(N ∩ σ). So α < sup(M ∩ η∗) < sup(N ∩ σ) = θ. Hence α ∈ Aη∗,γ ∩ θ = Aσ,γ ∩ θ, so α ∈ Aσ,γ . + Conversely, let α ∈ Aσ,γ ∩ M, and we will show that α ∈ Aη∗,γ . Since Q ∩ κ = + Aσ,γ , α ∈ Q ∩ M ∩ κ . We claim that α ∈ N. If N ∈ Y, then α ∈ Q ∈ N implies that α ∈ N. Suppose that N ∈ X . Then M < N, Q ∈ N ∩ Y, and Q ∩ κ = γ ∈ M ∩ N ∩ κ, which implies by Lemma 8.7 that Q ∩ M ∩ κ+ ⊆ N. In particular, α ∈ N. Hence in either case,

α ∈ Aσ,γ ∩ N = Aη∗,γ ∩ N ∩ σ.

So α ∈ Aη∗,γ . + We have proven that Aη∗,γ ∩ M = Aσ,γ ∩ M = Q ∩ M ∩ κ . Since Aη∗,γ is ∗ in M, it follows that Aη∗,γ is closed under H by Lemma 7.10. In particular, + P ∩ κ = Aη∗,γ . So + + P ∩ M ∩ κ = Aη∗,γ ∩ M = Aσ,γ ∩ M = Q ∩ M ∩ κ , which proves (4). Next we claim that P ∩ κ = γ and sup(P ) = η∗, which proves (2). Since P , γ, and η∗ are in M, it suﬃces to show that M models these statements. Let α ∈ P ∩ M ∩ κ, and we will show that α < γ. Then α ∈ P ∩ M ∩ κ+ = Q ∩ M ∩ κ+. So α ∈ Q ∩ κ = γ. Conversely, let α ∈ M ∩ γ, and we will show that α ∈ P ∩ κ. Then α ∈ M ∩ γ = M ∩ Q ∩ κ ⊆ M ∩ Q ∩ κ+ = M ∩ P ∩ κ+. 64 THOMAS GILTON AND JOHN KRUEGER

So α ∈ P . Now Q ∩ N ∩ [sup(M ∩ η∗), σ) is nonempty by assumption, so fix τ in this intersection. Then + τ ∈ Q ∩ κ ∩ N = Aσ,γ ∩ N = Aη∗,γ ∩ N ∩ σ. + ∗ ∗ So τ ∈ Aη∗,γ = P ∩ κ . Hence P ∩ N ∩ [sup(M ∩ η ), η ) 6= ∅, which proves (3). By Lemma 7.30, it follows that η∗ is a limit point of P . Since sup(P ∩ κ+) = ∗ ∗ ∗ sup(Aη∗,γ ) ≤ η , we have that sup(P ) ≤ η . As η is a limit point of P , it follows that sup(P ) = η∗, finishing the proof of (2). In particular, as the ordinals P ∩ κ = Q ∩ κ and η∗ both have uncountable cofinality, it follows that P ∈ Y by Lemma 7.15, which proves (1). We are now ready to prove the main lemma on proxies. This lemma contains all the information about proxies that we will need for applications. Lemma 11.5 (Main proxy lemma). Let M ∈ X and N ∈ X ∪ Y, where N is simple. Assume that M < N in the case that N ∈ X , and sup(M ∩ N ∩ κ) < N ∩ κ ∗ + in the case that N ∈ Y. Let η ∈ RN (M). Suppose that M 0 ∈ N ∩ X and N 0 ∈ N ∩ (X ∪ Y), where N 0 is simple. Assume that M 0 < N 0 in the case that N 0 ∈ X , sup(M 0 ∩ N 0 ∩ κ) < N 0 ∩ κ in the case that N 0 ∈ Y, and N ∈ X iff N 0 ∈ X . Suppose that M ∩ N = M 0 ∩ N 0 and p(M,N) = p(M 0,N 0). Assume that P ∈ M ∩ Y, P ∩ κ ∈ M ∩ N ∩ κ, cf(P ∩ κ) > ω, and τ ∈ P ∩ N ∩ [sup(M ∩ η∗), η∗). Then: (1) There is Q ∈ N 0 ∩ Y such that Q ∩ κ = P ∩ κ and Q ∩ N ∩ κ+ = P ∩ N ∩ η∗; in particular, τ ∈ Q. (2) If τ ∈ N 0, then there is P 0 ∈ M 0 ∩Y such that P 0 ∩κ = P ∩κ, P 0 ∩N 0 ∩κ+ = Q ∩ N 0 ∩ κ+, and P 0 ∩ M 0 ∩ κ+ = Q ∩ M 0 ∩ κ+; in particular, τ ∈ P 0. (3) If N ∈ Y, then there is P 0 ∈ M 0 ∩ Y such that P 0 ∩ κ = P ∩ κ and τ ∈ P 0. Moreover, if S~ is given and P is S~-strong, then the models Q and P 0 described in (1), (2), and (3) are also S~-strong.

+ ∗ Proof. Let k be the size of RN (M), and ﬁx i < k such that η is the i-th member + 0 of RN (M). Let a := p(M,N)(i) and a := {σ : ∃β (β, σ) ∈ a}.

(1) Let σ be the least ordinal in a0 such that P ∩ N ∩ η∗ ⊆ σ, which exists by Lemma 11.1(4). By Lemma 11.1(2(d)), min(a0) = min((N ∩ κ+) \ sup(M ∩ η∗)), which is strictly less than σ since P ∩N ∩[sup(M ∩η∗), η∗) 6= ∅ and P ∩N ∩η∗ ⊆ σ. Fix β such that (β, σ) ∈ a. By Lemma 11.1(5(a)), β ≤ P ∩ κ. We apply Lemma 11.3. Note that all of the assumptions of this lemma are satisﬁed. Let Q := Sk(Aσ,P ∩κ). Then by Lemma 11.3, (a) Q ∈ N ∩ Y; + (b) Q ∩ κ = P ∩ κ and Q ∩ κ = Aσ,P ∩κ; (c) Q ∩ N ∩ κ+ = P ∩ N ∩ η∗; (d) Q ∩ sup(N ∩ σ) = P ∩ sup(N ∩ σ). Since p(M,N) = p(M 0,N 0), it follows that p(M,N) ∈ N 0, and so in particular, σ ∈ N 0. And Q ∩ κ = P ∩ κ ∈ M ∩ N ∩ κ = M 0 ∩ N 0 ∩ κ ⊆ N 0. So σ and Q ∩ κ are 0 0 in N . Therefore Aσ,Q∩κ and Q are in N by elementarity. Properties (a), (b), and (c) above imply (1). MITCHELL’S THEOREM REVISITED 65

0 0 0 ∗ (2) Assume that τ ∈ N . We apply Lemma 11.4 to M and N . Let η0 be the + 0 0 0 i-th member of RN 0 (M ). Let a0 := p(M ,N )(i). Note that a0 = p(M,N)(i) = a 0 0 and a0 = a . Let us check that the assumptions of Lemma 11.4 are satisfied. Since p(M,N) = 0 0 0 0 0 0 p(M ,N ), (β, σ) ∈ p(M ,N )(i) = a0, and min(a0) = min(a ) < σ. We know that Q ∈ N 0 ∩ Y, β ≤ P ∩ κ = Q ∩ κ, Q ∩ κ = P ∩ κ ∈ M ∩ N ∩ κ = M 0 ∩ N 0 ∩ κ, + cf(Q ∩ κ) = cf(P ∩ κ) > ω, and Q ∩ κ = Aσ,Q∩κ. It remains to show that 0 0 ∗ Q ∩ N ∩ [sup(M ∩ η0 ), σ) 6= ∅. Since τ ∈ P ∩ N ∩ [sup(M ∩ η∗), η∗) and P ∩ N ∩ η∗ ⊆ N ∩ σ ⊆ sup(N ∩ σ), it follows that τ < σ, and τ ∈ P ∩ sup(N ∩ σ) = Q ∩ sup(N ∩ σ) by property (d) above. Also τ ∈ N 0 by assumption. So τ ∈ Q ∩ N 0 ∩ [sup(M ∩ η∗), η∗), and therefore min(a0) = min((N ∩ κ+) \ sup(M ∩ η∗)) ≤ τ. But then 0 ∗ 0 + 0 ∗ 0 0 sup(M ∩ η0 ) ≤ min((N ∩ κ ) \ sup(M ∩ η0 )) = min(a0) = min(a ) ≤ τ. 0 0 ∗ So τ ∈ Q∩N ∩[sup(M ∩η0 ), σ). This completes the verification of the assumptions of Lemma 11.4. 0 Let P := Sk(A ∗ ). Then by Lemma 11.4, η0 ,Q∩κ (i) P 0 ∈ M 0 ∩ Y; 0 0 + 0 ∗ (ii) P ∩ κ = Q ∩ κ, P ∩ κ = A ∗ , and sup(P ) = η ; η0 ,Q∩κ 0 0 0 0 ∗ ∗ (iii) P ∩ N ∩ [sup(M ∩ η0 ), η0 ) 6= ∅; (iv) P 0 ∩ M 0 ∩ κ+ = Q ∩ M 0 ∩ κ+. In particular, P 0 ∩ κ = P ∩ κ. It remains to prove that P 0 ∩ N 0 ∩ κ+ = Q ∩ N 0 ∩ κ+. We apply Lemma 11.3 to M 0, N 0, and P 0. Note that the assumptions of Lemma 0 11.3 are obviously satisfied, except for the claim that σ is the least ordinal in a0 such 0 0 ∗ 0 0 0 0 ∗ 0 that P ∩N ∩η0 ⊆ σ. So let σ be the least ordinal in a0 such that P ∩N ∩η0 ⊆ σ . 0 0 Then by Lemma 11.1(4(b)), σ is the largest ordinal in a0 such that for some γ, 0 0 0 0 γ ≤ P ∩ κ and (γ, σ ) ∈ a0. Now σ is the least ordinal in a = a0 such that ∗ 0 0 N ∩ P ∩ η ⊆ σ, so again by Lemma 11.1(4(b)), σ is the largest ordinal in a = a0 0 0 such that for some γ, γ ≤ P ∩ κ = P ∩ κ and (γ, σ) ∈ a = a0. So σ and σ satisfy 0 0 the same definition, and hence σ = σ . So indeed σ is the least ordinal in a0 such 0 0 ∗ that P ∩ N ∩ η0 ⊆ σ. 0 0 Since σ = σ and P ∩ κ = P ∩ κ, Q = Sk(Aσ,P ∩κ) = Sk(Aσ0,P 0∩κ). By 0 + 0 0 ∗ 0 Lemma 11.3, Q ∩ N ∩ κ = P ∩ N ∩ η . But P = Sk(A ∗ ), and therefore 0 η0 ,Q∩κ 0 ∗ 0 + 0 + 0 0 + P ∩ η0 = P ∩ κ . So Q ∩ N ∩ κ = P ∩ N ∩ κ . (3) If N ∈ Y, then N 0 ∈ Y. So Q ∈ N 0 implies that Q ⊆ N 0. Hence τ ∈ N 0. So we are done by (2).

Finally, the last statement follows from the properties of Q and P 0 described in (1) and (2) together with Lemma 5.6. 66 THOMAS GILTON AND JOHN KRUEGER

The main proxy lemma was concerned with the case that P ∈ M ∩ Y, P ∩ κ ∈ ∗ + ∗ ∗ M ∩ N ∩ κ, η ∈ RN (M), and τ ∈ P ∩ N ∩ [sup(M ∩ η ), η ). Another case which often occurs in the same contexts is that P ∈ M ∩ Y, P ∩ κ ∈ M ∩ N ∩ κ, and τ ∈ P ∩ N ∩ αM,N . This situation is handled by the next two lemmas. Lemma 11.6. Let M ∈ X and N ∈ X ∪ Y, where N is simple. Assume that M < N in the case that N ∈ X , and sup(M ∩ N ∩ κ) < N ∩ κ in the case that N ∈ Y. Suppose that M 0 ∈ N ∩ X and N 0 ∈ N ∩ (X ∪ Y), where N 0 is simple. Assume that M 0 < N 0 in the case that N 0 ∈ X , sup(M 0 ∩ N 0 ∩ κ) < N 0 ∩ κ in the case that N 0 ∈ Y, and N ∈ X iﬀ N 0 ∈ X . Suppose that M ∩ N = M 0 ∩ N 0 and p(M,N) = p(M 0,N 0). Assume that P ∈ M ∩ Y, P ∩ κ ∈ M ∩ N ∩ κ, cf(P ∩ κ) > ω, P ∩ αM,N is ∗ + unbounded in αM,N , and τ0 ∈ P ∩ N ∩ αM,N . Let η := min((M ∩ κ ) \ αM,N ). Then: (1) There is Q ∈ N 0 ∩ Y such that Q ∩ κ = P ∩ κ and Q ∩ N ∩ κ+ = P ∩ N ∩ η∗; in particular, τ0 ∈ Q. 0 0 0 0 0 (2) If τ0 ∈ N , then there is P ∈ M ∩ Y such that P ∩ κ = P ∩ κ and τ0 ∈ P . Moreover, if S~ is given and P is S~-strong, then the models Q and P 0 described in (1) and (2) are also S~-strong.

Proof. Since αM,N ∈ N by Lemma 8.2, αM,N ∈/ M. But αM,N ≤ sup(P ) and ∗ sup(P ) ∈ M. It follows that αM,N < sup(P ). Consequently, the ordinal η = + ∗ + min((M ∩ κ ) \ αM,N ) exists and is greater than αM,N . Therefore η ∈ RN (M). Since αM,N is a limit point of the countable set M ∩N, it follows that cf(αM,N ) = ω. As cf(P ∩ κ) > ω, we have that αM,N ∈ P by Lemma 7.14. So αM,N ∈ N ∩ P ∩ [sup(M ∩ η∗), η∗). We apply the main proxy lemma, Lemma 11.5, letting τ = αM,N . Then the ﬁrst statement of (1) above follows from Lemma 11.5(1). Since τ0 < αM,N , τ0 ∈ N ∩ P ∩ η∗ ⊆ Q. For (2), we have that 0 0 αM,N = sup(M ∩ N) = sup(M ∩ N ), which is in N 0 by Lemma 8.2. By Lemma 11.5(2), there is P 0 ∈ M 0 ∩ Y such that 0 0 0 + 0 + 0 P ∩ κ = P ∩ κ and P ∩ N ∩ κ = Q ∩ N ∩ κ . Assume that τ0 ∈ N . Then 0 + 0 0 + 0 τ0 ∈ Q ∩ N ∩ κ = P ∩ N ∩ κ ⊆ P . 0 Therefore τ0 ∈ P . Lemma 11.7. Let M ∈ X and N ∈ X ∪ Y, where N is simple. Assume that M < N in the case that N ∈ X , and sup(M ∩ N ∩ κ) < N ∩ κ in the case that N ∈ Y. Suppose that P ∈ M ∩ Y, P ∩ κ ∈ M ∩ N ∩ κ, P ∩ αM,N is bounded below αM,N , 0 0 and τ ∈ P ∩ N ∩ αM,N . Then there is P ∈ M ∩ N ∩ Y such that P ∩ κ = P ∩ κ and τ ∈ P 0. Moreover, if S~ is given and P is S~-strong, then P 0 is S~-strong.

Proof. Let α := αM,N and δ := sup(M ∩ N ∩ κ). Deﬁne

σ := sup(P ∩ Aα,δ). By Lemma 8.11, σ satisﬁes: (a) σ ∈ M ∩ N ∩ κ+; (b) P ∩ σ = Aσ,P ∩κ; MITCHELL’S THEOREM REVISITED 67

+ (c) P ∩ (M ∩ N) ∩ κ = Aσ,P ∩κ ∩ (M ∩ N); (d) N ∩ P ∩ αM,N ⊆ Aσ,P ∩κ.

Since σ and P ∩ κ are in M ∩ N, Lemma 7.10 and (c) imply that Aσ,P ∩κ is closed ∗ 0 0 under H . Let P := Sk(Aσ,P ∩κ). Then P is in M ∩ N. By (b), we have that 0 P ∩ κ = Aσ,P ∩κ ∩ κ = P ∩ σ ∩ κ = P ∩ κ. 0 0 By (d), τ ∈ Aσ,P ∩κ ⊆ P . It remains to show that P ∈ Y. It suffices to show that 0 0 lim(Csup(P 0)) ∩ P is cofinal in sup(P ). Since P and Aα,δ are closed under successors, P ∩ Aα,δ has no maximal element. As σ is a limit point of P , 0 sup(P ) = sup(Aσ,P ∩κ) = sup(P ∩ σ) = σ. 0 + Also P ∩ κ = Aσ,P ∩κ = P ∩ σ. So it is enough to show that lim(Cσ) ∩ P is cofinal in σ. Now σ is a limit point of P , and therefore has cofinality less than κ. If σ∈ / P , then σ ∈ cl(P ) \ P , so by Lemma 7.13, lim(Cσ) ∩ P is cofinal in σ and we are done. Otherwise σ ∈ P . By the definition of σ, σ is not in Aα,δ. Now Aα,δ is closed under ∗ H by Lemma 7.29. So Q := Sk(Aα,δ) is an elementary substructure of A with + 0 + 0 Q ∩ κ = Aα,δ. Let σ := min((Q ∩ κ ) \ σ), which exists since σ < α. Then σ 0 has uncountable cofinality, which implies that lim(Cσ0 ) is cofinal in σ . Also by the elementarity of Q, σ is a limit point of Cσ0 , and therefore

Cσ = Cσ0 ∩ σ. 0 Again by the elementarity of Q, lim(Cσ0 ) ∩ Q is cofinal in sup(Q ∩ σ ) = σ. In particular, lim(Cσ) is cofinal in σ. Since σ ∈ P and σ has cofinality less than κ, ot(Cσ) ∈ P ∩ κ, and therefore Cσ ⊆ P . Hence lim(Cσ) ∩ P = lim(Cσ), and this set is cofinal in σ as observed above. Finally, assume that P is S~-strong. Then P 0 ∩ κ = P ∩ κ, P 0 ∈ M ∩ N, and by (b), 0 + P ∩ (M ∩ N) ∩ κ = Aσ,P ∩κ ∩ (M ∩ N) ⊆ Aσ,P ∩κ ⊆ P. 0 ~ So P is S-strong by Lemma 5.6.

§12. The proxy construction

Let M ∈ X and N ∈ X ∪Y, where N is simple. Assume that M < N in the case ∗ + that N ∈ X , and sup(M ∩N ∩κ) < N ∩κ in the case that N ∈ Y. Let η ∈ RN (M). We will prove that there exist sets a and a0 satisfying properties (1)–(5) of Lemma 11.1.8 We recall a well-ordering on finite sets of ordinals which was used in [12]. For finite sets of ordinals x and y, define x ≺ y if x 6= y and max(x4y) ∈ y. Lemma 12.1. The relation ≺ is a well-ordering of [On]<ω.

8Our proof of the proxy existence lemma is based on the construction of Mitchell [12, Lemma 3.46]. We point out that there is a mistake in Mitchell’s construction. The problem arises in the case when the ordinal η from that proof is deﬁned as max(lim(Cα) ∩ X), and η happens to have dropped below sup(M 0). In this case, there appears to be no reason why recursion hypothesis (1c) can be maintained. This problem was discovered by Gilton, and later corrected by Krueger. 68 THOMAS GILTON AND JOHN KRUEGER

Proof. It is obvious that ≺ is irreflexive and total. For transitivity, let x ≺ y ≺ z, and we will show that x ≺ z. Let α := max(x4y), β := max(y4z), and γ := max(x4z). Then α ∈ y \ x and β ∈ z \ y. We will show that γ ∈ z. Suppose for a contradiction that γ∈ / z, so that γ ∈ x. The following statements can be easily proved: (1) α, β, and γ are distinct; (2) α ∈ z implies that α < γ; (3) α∈ / z implies that α < β; (4) β ∈ x implies that β < α; (5) β∈ / x implies that β < γ; (6) γ ∈ y implies that γ < β; and (7) γ∈ / y implies that γ < α. Now one can easily check by inspection that any Boolean combination of the statements α ∈ z, β ∈ x, and γ ∈ y yields a contradiction. For example, suppose that α ∈ z, β ∈ x, and γ ∈ y. Then (2), (4), and (6) imply that α < γ, β < α, and γ < β, which in turn imply that α < γ < β < α, which is absurd. The other possibilities are ruled out in a similar manner. This completes the proof that ≺ is transitive. To show that ≺ is a well-ordering, suppose for a contradiction that hxn : n < ωi is a ≺-decreasing sequence of finite sets of ordinals. We define by induction an increasing sequence hkn : n < ωi of integers and a ⊆-decreasing sequence hAn : n < ωi of infinite subsets of ω as follows. Let k0 = 0 and A0 = ω. Assume that kn and An are defined, where An is an infinite subset of ω. Let kn+1 be the least integer in An strictly greater than kn. Now for all r ∈ An with r > kn+1, we have that xr ≺ xkn+1 , and hence max(xr4xkn+1 ) ∈ xkn+1 . Since xkn+1 is finite and An is infinite, we can find an infinite subset An+1 of An \ (kn+1 + 1) such that for all r, s ∈ An+1, max(xr4xkn+1 ) = max(xs4xkn+1 ).

This completes the construction. For each n, let αn := max(xkn 4xkn+1 ). We claim that hαn : n < ωi is a descending sequence of ordinals, which gives a contradiction. Let n < ω. Since xkn+1 ≺ xkn , αn ∈ xkn \ xkn+1 . So clearly αn 6= αn+1. Suppose for a contradiction that αn < αn+1. Then by the maximality of αn, αn+1 cannot be in xkn 4xkn+1 , and therefore must be in xkn ∩xkn+1 . But by construction, max(xkn 4xkn+2 ) = αn. Therefore αn+1 must be in xkn+2 , since otherwise it is in xkn 4xkn+2 but larger than αn. This contradicts that αn+1 = max(xkn+1 4xkn+2 ) is in xkn+1 \ xkn+2 .

We will define by induction two sequences of sets a0, . . . , an and b0, . . . , bn. The induction stops when bn = ∅. Each ak and bk will be a finite set of pairs of ordinals. 0 0 We let ak := {σ : ∃β (β, σ) ∈ ak} and bk := {η : ∃β (β, η) ∈ bk}. 0 0 0 By construction, for each k, bk+1 will be equal to (bk \{η})∪x, where η = min(bk) 0 0 and x is a finite subset of η. In particular, max(bk4bk+1) will be equal to η, which 0 0 0 0 is in bk, and hence bk+1 ≺ bk. Therefore the sequence of bk’s is ≺-descending, and so must terminate with the empty set after finitely many steps. When defining these sequences, we will maintain the following inductive hypotheses:

+ (A) For all (β, σ) ∈ ak, β ∈ M ∩ N ∩ κ, σ ∈ N ∩ κ is a limit ordinal, and ∗ 0 sup(N ∩ σ) ≤ η . The least member of ak, if it exists, is equal to min((N ∩ + ∗ 0 κ ) \ sup(M ∩ η )). For each σ ∈ ak, there is a unique β with (β, σ) ∈ ak.

∗ (B) For all (β, η) ∈ bk, β ∈ M ∩ N ∩ κ and η ≤ η is a limit ordinal. If η0 < η1 0 0 are successive elements of bk, then N ∩ [η0, η1) 6= ∅. For each η ∈ bk, there is a unique β with (β, η) ∈ bk. MITCHELL’S THEOREM REVISITED 69

0 0 (C) If bk 6= ∅, then ak 6= ∅ and max(ak) < min(bk).

0 (D) If (β, σ) ∈ ak and min(ak) < σ, then for all γ with β ≤ γ < κ,

Aη∗,γ ∩ sup(N ∩ σ) = Aσ,γ ∩ sup(N ∩ σ).

(E) If (β, η) ∈ bk, then for all γ with β ≤ γ < κ,

Aη,γ = Aη∗,γ ∩ η.

(F) If (β, η) ∈ bk, then whenever P ∈ M ∩ Y is such that P ∩ κ ∈ M ∩ N ∩ κ and P ∩ N ∩ [η−, η) 6= ∅, − 0 0 where η is the largest ordinal in ak ∪ bk less than η, then β ≤ P ∩ κ. (G) Whenever P ∈ M ∩ Y is such that P ∩ κ ∈ M ∩ N ∩ κ and P ∩ N ∩ [sup(M ∩ η∗), η∗) 6= ∅, ∗ 0 0 then P ∩ N ∩ η ⊆ max(ak ∪ bk). (H) Suppose that P ∈ M ∩ Y and τ satisfy that P ∩ κ ∈ M ∩ N ∩ κ and τ ∈ P ∩ N ∩ [sup(M ∩ η∗), η∗). 0 0 Let σ := min((ak ∪ bk) \ (τ + 1)), which exists by (G), and assume that 0 σ ∈ ak. Fix β with (β, σ) ∈ ak. Then: (i) β ≤ P ∩ κ; (ii) P ∩ sup(N ∩ σ) = Aσ,P ∩κ ∩ sup(N ∩ σ).

0 (I) If (β, σ) ∈ ak ∪ bk, where min(ak) < σ, then P := Sk(Aη∗,β) satisﬁes that + − ∗ P ∈ M ∩ Y,P ∩ κ = β, P ∩ κ = Aη∗,β, and P ∩ N ∩ [σ , η ) 6= ∅, − 0 0 where σ is the largest member of ak ∪ bk less than σ.

∗ Note that since αM,N is a limit point of M and αM,N < η , it follows that ∗ αM,N ≤ sup(M ∩ η ). 0 0 Suppose that P is as in (G), and σ is the least ordinal in ak ∪ bk such that P ∩ N ∩ η∗ ⊆ σ. By the minimality of σ, we can ﬁx τ ∈ P ∩ N ∩ η∗ such that − − 0 0 σ ≤ τ, where σ is the greatest member of ak ∪ bk less than σ. Since N and P are closed under successors, τ + 1 ∈ P ∩ N ∩ η∗. As P ∩ N ∩ η∗ ⊆ σ, it follows that 0 0 σ = min((ak ∪ bk) \ (τ + 1)). In the arguments which follow, we will frequently consider models P ∈ M ∩ Y such that P ∩ [sup(M ∩ η∗), η∗) 6= ∅, for example, in (G) and (H). Note that by Lemma 7.30, for any such P , η∗ is a limit point of P . Therefore by Lemma 7.27, ∗ P ∩ η = Aη∗,P ∩κ.

Assume that a0, . . . , an and b0, . . . , bn are sequences satisfying properties (A)– (I), where n is the least integer such that bn = ∅. Let us show that the sets a := an and a0 := {σ : ∃β (β, σ) ∈ a} satisfy properties (1)–(5) in the conclusion of Lemma 11.1. (1) is immediate, and (2) follows from (A). (3(a)) follows from (D). For (3(b)), let us prove that the equation

Aη∗,γ ∩ N ∩ σ = Aσ,γ ∩ N 70 THOMAS GILTON AND JOHN KRUEGER follows from the equation

Aη∗,γ ∩ sup(N ∩ σ) = Aσ,γ ∩ sup(N ∩ σ), which holds by (3(a)). Let ξ ∈ Aη∗,γ ∩N ∩σ, and we will show that ξ ∈ Aσ,γ . Then ξ < sup(N ∩ σ), so by the last equation, ξ ∈ Aσ,γ . Conversely, let ξ ∈ Aσ,γ ∩ N, and we will show that ξ ∈ Aη∗,γ . Then ξ ∈ N ∩ σ, so ξ < sup(N ∩ σ). By the last equation, ξ ∈ Aη∗,γ . (4) Suppose that P ∈ M ∩ Y,P ∩ κ ∈ M ∩ N ∩ κ, and P ∩ N ∩ [sup(M ∩ η∗), η∗) 6= ∅. 0 By (G) and the fact that bn = ∅, P ∩ N ∩ η∗ ⊆ max(a0). Let σ ∈ a0 be the least ordinal such that P ∩N ∩η∗ ⊆ σ. Fix β such that (β, σ) ∈ a. Define X := {(β0, σ0) ∈ a : β0 ≤ P ∩ κ} and X0 := {σ0 : ∃β0 (β0, σ0) ∈ X}. We will prove that σ = max(X0), which completes the proof of (4). By the minimality of σ, clearly there is τ ∈ N ∩ P ∩ [sup(M ∩ η∗), η∗) such that σ = min(a0 \(τ +1)). By (H), β ≤ P ∩κ. It follows that (β, σ) ∈ X, and so σ ∈ X0. Suppose for a contradiction that there is σ0 ∈ X0 which is larger than σ. Fix β0 with (β0, σ0) ∈ X. Then σ ≤ (σ0)−, where (σ0)− is the largest member of a0 which is less than σ0. By (I), 0 − ∗ Aη∗,β0 ∩ N ∩ [(σ ) , η ) 6= ∅. Since σ ≤ (σ0)−, it follows that ∗ Aη∗,β0 ∩ N ∩ [σ, η ) 6= ∅. As β0 ≤ P ∩ κ by the definition of X, ∗ Aη∗,β0 ⊆ Aη∗,P ∩κ = P ∩ η . But then P ∩ N ∩ [σ, η∗) 6= ∅, which contradicts that P ∩ N ∩ η∗ ⊆ σ. (5) Suppose that P ∈ M ∩ Y satisfies that P ∩ κ ∈ M ∩ N ∩ κ and P ∩ N ∩ [sup(M ∩ η∗), η∗) 6= ∅, σ is the least ordinal in a0 such that P ∩ N ∩ η∗ ⊆ σ, and (β, σ) ∈ a. By the minimality of σ, we can fix τ ∈ P ∩ N ∩ [sup(M ∩ η∗), η∗) such that σ = min(a0 \ (τ + 1)). Then (5(a,b)) follow immediately from (H(i,ii)). For (5(c)), P ∩ N ∩ η∗ = P ∩ N ∩ σ, and

P ∩ N ∩ σ = P ∩ sup(N ∩ σ) ∩ N = Aσ,P ∩κ ∩ sup(N ∩ σ) ∩ N = Aσ,P ∩κ ∩ N.

We now turn to proving that there exist sequences a0, . . . , an and b0, . . . , bn satisfying properties (A)–(I), where n is the least integer such that bn = ∅.

First we consider the base case. Let β be the least ordinal in M ∩ N ∩ κ for which there exists P ∈ M ∩ Y such that P ∩ κ = β and P ∩ N ∩ [sup(M ∩ η∗), η∗) 6= ∅.

If there is no such β, then let a0 = ∅ and b0 = ∅, and we are done. Suppose that β exists. Then obviously N ∩ [sup(M ∩ η∗), η∗) 6= ∅. Deﬁne + ∗ ∗ a0 := {(0, min((N ∩ κ ) \ sup(M ∩ η )))} and b0 := (β, η ). MITCHELL’S THEOREM REVISITED 71

In the case that a0 = b0 = ∅, the inductive hypotheses are all vacuously true. In the other case, the inductive hypotheses are all either vacuously true or trivial.

Now we handle the induction step. Assume that k < ω and ak and bk have been deﬁned and satisfy the inductive hypotheses. If bk = ∅, then we are done. Assume 0 that bk is nonempty. Then by (C), ak 6= ∅. Let η be the least member of bk, and 0 let β be the unique ordinal such that (β, η) ∈ bk. By (A) and (B), max(ak) and η 0 are limit ordinals. By (C), max(ak) < η, and in particular, ω < η.

First consider the easy case that η = η0 +ω for some limit ordinal η0. Let ak+1 := 0 ak. If max(ak) < η0, then let bk+1 := (bk \{(β, η)}) ∪ {(β, η0)}. Suppose that η0 ≤ 0 0 0 0 max(ak). Since max(ak) is a limit ordinal and max(ak) < η, clearly max(ak) = η0. In this case, let bk+1 := bk \{(β, η)}. All of the inductive hypotheses can be easily checked, using Notation 7.2(4) and the fact that if P ∈ Y and P ∩ N ∩ [η0, η) 6= ∅, then by the elementarity of P ∩ N, η ∈ P ∩ N.

From now on we will assume that η is a limit of limit ordinals. In particular, every ordinal in Cη is a limit ordinal by Notation 7.2(3).

Deﬁne

θ := sup(lim(Cη) ∩ cl(N)).

We split the deﬁnition of ak+1 and bk+1 into two cases.

Case 1: θ = sup(N).

0 0 Note that since max(ak) ∈ N and sup(N) = θ, it follows that max(ak) < θ.

0 0 0 Claim 1: η = max(bk). Suppose for a contradiction that there is η ∈ bk greater 0 0 0 0 − 0 − than η. Fix β with (β , η ) ∈ bk. Then η ≤ (η ) , where (η ) is the largest ordinal 0 0 0 0 − ∗ in ak ∪bk less than η . By (I), Aη∗,β0 ∩N ∩[(η ) , η ) 6= ∅. Fix τ in this intersection. Then θ ≤ η ≤ (η0)− ≤ τ and τ ∈ N, which contradicts that θ = sup(N).

Since N is simple, ot(Cθ) = sup(N ∩ κ). Let ξ := sup(M ∩ N ∩ κ), which is in N ∩ κ by Lemma 1.30. Define σ := cθ,ξ. Since θ = sup(N) and ξ ∈ N, σ ∈ N by Lemma 7.19. As ξ has countable cofinality, so does σ. In particular, N ∩ σ is cofinal in σ.

Claim 2: Aθ,ξ = Aσ,ξ and sup(Aσ,ξ) = σ. As ξ is a limit ordinal, σ = cθ,ξ ∈ lim(Cθ). Therefore Aσ,ξ = Aθ,ξ ∩ σ. In particular, Aσ,ξ ⊆ Aθ,ξ. On the other hand, since ξ < sup(N ∩ κ) = ot(Cθ), it follows that Aθ,ξ ⊆ cθ,ξ = σ by Notation 7.4(6). So Aθ,ξ ⊆ Aθ,ξ ∩ σ = Aσ,ξ. This proves that Aθ,ξ = Aσ,ξ. Since ξ = sup(M ∩ N ∩ κ), by the elementarity of M ∩ N, ξ is a limit of limit ordinals. Therefore σ = cθ,ξ is a limit of lim(Cθ) ∩ σ. For any ordinal ζ ∈ lim(Cθ)∩σ, the fact that ζ ∈ lim(Cθ)∩cθ,ξ and ξ < ot(Cθ) implies by Notation 7.4(6) that ζ ∈ Aθ,ξ. So lim(Cθ) ∩ σ is coﬁnal in σ and is a subset of Aθ,ξ. Hence sup(Aσ,ξ) = sup(Aθ,ξ) = σ.

0 We now deﬁne ak+1 and bk+1. Let bk+1 := ∅. If σ ≤ max(ak), then let ak+1 := 0 ak. If max(ak) < σ, then let ak+1 := ak ∪ {(β, σ)}. 72 THOMAS GILTON AND JOHN KRUEGER

We prove that the inductive hypotheses are maintained. First, consider the case 0 when σ ≤ max(ak). Then ak+1 = ak and bk+1 = ∅. Inductive hypotheses (A), (B), (C), (D), (E), (F), and (I) are all either vacuously true, or follow immediately from the inductive hypotheses. For (G) and (H), suppose that P ∈ M ∩ Y satisﬁes that P ∩ κ ∈ M ∩ N ∩ κ and P ∩ N ∩ [sup(M ∩ η∗), η∗) 6= ∅. By inductive hypothesis (G) and Claim 1, ∗ 0 0 P ∩ N ∩ η ⊆ max(ak ∪ bk) = η. ∗ 0 If P ∩N ∩η ⊆ max(ak), then this proves (G) for k +1, and in that case (H) follows immediately from the inductive hypotheses. 0 Otherwise P ∩ N ∩ [max(ak), η) 6= ∅. Let us show that this is impossible. Since 0 0 0 max(ak) is the predecessor of η in ak ∪ bk, inductive hypothesis (F) implies that β ≤ P ∩ κ. Inductive hypothesis (E) then implies that

Aη,P ∩κ = Aη∗,P ∩κ ∩ η. As η∗ is a limit point of P ,

P ∩ η = Aη∗,P ∩κ ∩ η = Aη,P ∩κ.

Since θ ∈ lim(Cη), Aθ,P ∩κ = Aη,P ∩κ ∩ θ = P ∩ θ. As θ = sup(N), P ∩ N ∩ η = P ∩ N ∩ θ = Aθ,P ∩κ ∩ N. And as P ∩ κ ∈ M ∩ N ∩ κ and ξ = sup(M ∩ N ∩ κ), it follows that P ∩ κ ≤ ξ, so

Aθ,P ∩κ ⊆ Aθ,ξ = Aσ,ξ by Claim 2. So 0 P ∩ N ∩ η = Aθ,P ∩κ ∩ N ⊆ Aσ,ξ ⊆ σ ≤ max(ak). 0 This contradicts the initial assumption that P ∩ N ∩ [max(ak), η) 6= ∅.

0 Secondly, consider the case that max(ak) < σ. Then ak+1 = ak ∪ {(β, σ)} and bk+1 = ∅. We prove that the inductive hypotheses are maintained. Inductive hypotheses (A), (B), (C), (E), and (F) are all either vacuously true, or follow immediately from the inductive hypotheses. It remains to show (D), (G), (H), and (I).

(D) By inductive hypothesis (D), we only need to check that (D) holds for k + 1 in the case of (β, σ). As noted in the paragraph before Claim 2, N ∩ σ is coﬁnal in σ. Also observe that since θ ∈ lim(Cη), σ = cθ,ξ = cη,ξ is a limit point of Cη. Let β ≤ γ < κ be given. Since (β, η) ∈ bk, inductive hypothesis (E) implies that Aη,γ = Aη∗,γ ∩ η. Since σ is a limit point of Cη and sup(N ∩ σ) = σ, it follows that Aσ,γ ∩ sup(N ∩ σ) = Aσ,γ ∩ σ = Aσ,γ = Aη,γ ∩ σ = (Aη∗,γ ∩ η) ∩ σ = Aη∗,γ ∩ σ = Aη∗,γ ∩ sup(N ∩ σ), which proves (D). (G) Suppose that P ∈ M ∩ Y satisﬁes that P ∩ κ ∈ M ∩ N ∩ κ and P ∩ N ∩ [sup(M ∩ η∗), η∗) 6= ∅. By inductive hypothesis (G) and Claim 1, ∗ 0 0 P ∩ N ∩ η ⊆ max(ak ∪ bk) = η. MITCHELL’S THEOREM REVISITED 73

∗ 0 0 0 If P ∩ N ∩ η ⊆ max(ak), then since max(ak) ≤ max(ak+1), we are done. ∗ 0 So assume that there exists τ ∈ (P ∩ N ∩ η ) \ max(ak). We will show that ∗ 0 ∗ P ∩N ∩η ⊆ σ, which completes the proof since σ = max(ak+1). As P ∩N ∩η ⊆ η, 0 0 it follows that τ ∈ P ∩ N ∩ [max(ak), η). Since max(ak) is the largest member of 0 0 ak ∪ bk less than η, inductive hypothesis (F) implies that β ≤ P ∩ κ. Inductive hypothesis (E) then implies that

Aη,P ∩κ = Aη∗,P ∩κ ∩ η.

But θ ∈ lim(Cη) and θ = sup(N), so

P ∩ N ∩ η = Aη∗,P ∩κ ∩ N ∩ η = Aη,P ∩κ ∩ N = Aη,P ∩κ ∩ N ∩ θ = Aθ,P ∩κ ∩ N.

Hence P ∩ N ∩ η ⊆ Aθ,P ∩κ. Since P ∩ κ ∈ M ∩ N ∩ κ, P ∩ κ < sup(M ∩ N ∩ κ) = ξ. So by Claim 2, P ∩ N ∩ η ⊆ Aθ,P ∩κ ⊆ Aθ,ξ = Aσ,ξ ⊆ σ. Since P ∩ N ∩ η∗ ⊆ η as noted above, we have that ∗ 0 P ∩ N ∩ η = P ∩ N ∩ η ⊆ σ = max(ak+1).

(H) Suppose that P ∈ M ∩ Y and τ satisfy that P ∩ κ ∈ M ∩ N ∩ κ and τ ∈ P ∩ N ∩ [sup(M ∩ η∗), η∗). 0 0 0 0 0 0 Let σ = min(ak+1 \ (τ + 1)). Fix β with (β , σ ) ∈ ak+1. If σ < σ, then clearly 0 0 σ = min(ak \ (τ + 1)), so (i) and (ii) follow from inductive hypothesis (H). 0 0 0 Suppose that σ = σ, which means that max(ak) ≤ τ + 1. Then β = β. Since 0 0 0 max(ak) is the largest ordinal in ak ∪bk less than η, inductive hypothesis (F) implies that β ≤ P ∩ κ, which proves (i). By inductive hypothesis (E),

Aη,P ∩κ = Aη∗,P ∩κ ∩ η.

Since σ ∈ lim(Cη),

P ∩ σ = Aη∗,P ∩κ ∩ σ = Aη,P ∩κ ∩ σ = Aσ,P ∩κ = Aσ,P ∩κ ∩ σ.

As σ = sup(N ∩ σ), we have that P ∩ sup(N ∩ σ) = Aσ,P ∩κ ∩ sup(N ∩ σ), which proves (ii).

(I) By inductive hypothesis (I), it suﬃces to consider (β, σ). Let P := Sk(Aη∗,β). 0 0 0 Since (β, η) ∈ bk and max(ak) is the largest ordinal in ak ∪ bk less than η, inductive + hypothesis (I) implies that P ∈ M ∩ Y, P ∩ κ = β, P ∩ κ = Aη∗,β, and P ∩ N ∩ 0 ∗ 0 0 0 [max(ak), η ) 6= ∅. Since max(ak) is also the largest ordinal in ak+1 ∪bk+1 less than σ, we are done.

Case 2: θ < sup(N).

0 + 0 0 Let σ := min((N ∩ κ ) \ θ), which exists by Case 2. If σ ≤ max(ak), then 0 0 0 0 let σ := max(ak) and ak+1 := ak. If max(ak) < σ , then let σ := σ and ak+1 := ak ∪ {(β, σ)}.

Deﬁne A as the set of ordinals of the form min(Cη \ (ξ + 1)), where for some P ∈ M ∩ Y with P ∩ κ ∈ M ∩ N ∩ κ, ξ ∈ P ∩ N ∩ [σ, η).9

9The set A could be empty. In fact, it is possible for example that θ = η∗ and σ = min((N ∩ κ+) \ η∗), so that η < σ. In this case we interpret [σ, η) to be the empty set, so that A is empty as well. 74 THOMAS GILTON AND JOHN KRUEGER

Note that every ordinal in A is a limit ordinal, since Cη consists of limit ordinals, and is strictly greater than σ. Suppose that γ0 < γ1 are in A. Then for some ξ ∈ P ∩ N ∩ [σ, η), γ1 = min(Cη \ (ξ + 1)). So γ0 ≤ ξ < ξ + 1 < γ1. In particular, N ∩ [γ0, γ1) 6= ∅.

Claim: A is finite. Suppose for a contradiction that A is infinite. Fix an increasing sequence hγn : n < ωi from A. Then σ < γ0, and for each n, γn ∈ Cη and N ∩[γn, γn+1) 6= ∅. It follows that the ordinal sup{γn : n < ω} is in lim(Cη)∩cl(N), and yet is greater than σ and hence θ. This contradicts the definition of θ.

For each δ ∈ A, define βδ as the least ordinal in M ∩ N ∩ κ such that for some P ∈ M ∩ Y, P ∩ κ = βδ, and there is ξ ∈ P ∩ N ∩ [σ, η) such that δ = 0 min(Cη \ (ξ + 1)). Note that βδ exists by the definition of A. Also as max(ak) ≤ σ, 0 0 0 0 P ∩ N ∩ [max(ak), η) 6= ∅. Therefore since max(ak) is the largest ordinal in ak ∪ bk less than η, inductive hypothesis (F) implies that β ≤ P ∩ κ = βδ. Define bk+1 := (bk \{(β, η)}) ∪ {(βδ, δ): δ ∈ A}.

We verify the inductive hypotheses. Hypotheses (A) and (B) are straightforward to check.

0 (C) We know that ak+1 6= ∅ and max(ak+1) = σ. If A is nonempty, then 0 0 max(ak+1) = σ < min(A) = min(bk+1). 0 0 If A is empty and bk+1 is nonempty, then min(bk+1) is the least member of bk 0 0 greater than η. So if max(ak+1) = max(ak), then by inductive hypothesis (C), 0 0 0 0 max(ak+1) = max(ak) < min(bk) = η < min(bk+1). 0 Suppose that max(ak) < σ. By inductive hypothesis (B), we have that N ∩ 0 0 + [η, min(bk+1)) 6= ∅. Since θ ≤ η, N ∩ [θ, min(bk+1)) 6= ∅. As σ = min((N ∩ κ ) \ θ), 0 0 this implies that max(ak+1) = σ < min(bk+1).

(D) By inductive hypothesis (D), it suﬃces to consider (β, σ) in the case where 0 + max(ak) < σ and ak+1 = ak ∪ {(β, σ)}. So σ = min((N ∩ κ ) \ θ), and therefore θ = sup(N ∩ σ). Let β ≤ γ < κ. Then by Lemma 7.12(2),

Aθ,γ = Aσ,γ ∩ θ.

Since (β, η) ∈ bk, inductive hypothesis (E) implies that

Aη,γ = Aη∗,γ ∩ η.

And since θ ∈ lim(Cη), Aθ,γ = Aη,γ ∩ θ. Therefore

Aσ,γ ∩ θ = Aθ,γ = Aη,γ ∩ θ = Aη∗,γ ∩ θ. Since sup(N ∩ σ) = θ, this proves (D).

(E) Consider (βδ, δ) ∈ bk+1, where δ ∈ A. Fix P ∈ M ∩ Y and ξ such that P ∩ κ = βδ, ξ ∈ P ∩ N ∩ [σ, η), and δ = min(Cη \ (ξ + 1)). Let βδ ≤ γ < κ, and we will show that

Aδ,γ = Aη∗,γ ∩ δ. MITCHELL’S THEOREM REVISITED 75

As observed above, β ≤ βδ ≤ γ. Since (β, η) ∈ bk, by inductive hypothesis (E),

∗ ∗ Aη,βδ = Aη ,βδ ∩ η and Aη,γ = Aη ,γ ∩ η. As ξ ∈ N ∩ P , by elementarity ξ + 1 ∈ N ∩ P . Hence

∗ ∗ ξ + 1 ∈ P ∩ η = Aη ,P ∩κ ∩ η = Aη ,βδ ∩ η = Aη,βδ .

So ξ + 1 ∈ Aη,βδ \ Cη. By Notation 7.4(7), min(Cη \ (ξ + 1)) = δ is in Aη,βδ . Since βδ ≤ γ, δ ∈ Aη,γ . Therefore

Aδ,γ = Aη,γ ∩ δ = (Aη∗,γ ∩ η) ∩ δ = Aη∗,γ ∩ δ, proving (E).

(F) Let (γ, ζ) ∈ bk+1. Then either (γ, ζ) = (βδ, δ) for some δ ∈ A, or (γ, ζ) ∈ bk and η < ζ.

Case a: (γ, ζ) = (βδ, δ) for some δ ∈ A. Suppose that P ∈ M ∩ Y satisﬁes that P ∩ κ ∈ M ∩ N ∩ κ and P ∩ N ∩ [δ−, δ) 6= ∅, − 0 0 where δ is the greatest member of ak+1 ∪ bk+1 which is less than δ. Then clearly 0 − − σ = max(ak+1) ≤ δ . So if we ﬁx ξ ∈ P ∩ N ∩ [δ , δ), then σ ≤ ξ and δ = min(Cη \ (ξ + 1)). By the minimality of βδ, βδ ≤ P ∩ κ.

0 Case b: (γ, ζ) ∈ bk and η < ζ. If ζ is not the least element of bk greater than η, 0 0 then the greatest ordinal in ak ∪ bk less than ζ is equal to the greatest ordinal in 0 0 ak+1 ∪ bk+1 less than ζ. In that case, (F) follows easily from inductive hypothesis (F). 0 Suppose that ζ is the least member of bk greater than η. Then the greatest 0 0 − member of ak+1 ∪ bk+1 less than ζ, which we denote by ζ , is equal to either max(A) if A is nonempty, or σ if A is empty. Assume that P ∈ M ∩ Y satisﬁes that P ∩ κ ∈ M ∩ N ∩ κ and P ∩ N ∩ [ζ−, ζ) 6= ∅. We will show that γ ≤ P ∩ κ. If P ∩ N ∩ [η, ζ) 6= ∅, then since η is the greatest 0 0 member of ak ∪ bk less than ζ, γ ≤ P ∩ κ by inductive hypothesis (F). Otherwise P ∩N ∩[ζ−, η) 6= ∅. Fix ξ in this intersection. Then ξ+1 is also in this intersection, by the elementarity of P ∩ N and because η is a limit ordinal. By the − deﬁnition of A, min(Cη \(ξ+1)) is in A. So A is nonempty, and hence ζ = max(A). − Yet min(Cη \ (ξ + 1)) is in A and is strictly greater than ζ = max(A), which is a contradiction.

(G) Let P ∈ M ∩ Y satisfy that P ∩ κ ∈ M ∩ N ∩ κ and P ∩ N ∩ [sup(M ∩ η∗), η∗) 6= ∅. By inductive hypothesis (G), ∗ 0 0 0 P ∩ N ∩ η ⊆ max(ak ∪ bk) = max(bk). 0 0 0 If max(bk) = max(bk+1), then we are done. Otherwise η is equal to max(bk), so ∗ ∗ 0 0 P ∩ N ∩ η ⊆ η. If P ∩ N ∩ η is not a subset of max(ak+1 ∪ bk+1), then there is ∗ 0 0 ξ ∈ P ∩ N ∩ η such that max(ak+1) = σ ≤ ξ, and also max(A) = max(bk+1) ≤ ξ if A is nonempty. So ξ ∈ P ∩ N ∩ [σ, η), which implies that min(Cη \ (ξ + 1)) is in A. So A is nonempty, and max(A) ≤ ξ < min(Cη \ (ξ + 1)) ∈ A, which is impossible. 76 THOMAS GILTON AND JOHN KRUEGER

(H) Suppose that P ∈ M ∩ Y and τ satisfy P ∩ κ ∈ M ∩ N ∩ κ and τ ∈ P ∩ N ∩ [sup(M ∩ η∗), η∗). Let ∗ 0 0 σ := min((ak+1 ∪ bk+1) \ (τ + 1)), ∗ 0 ∗ ∗ ∗ and assume that σ ∈ ak+1. Fix β with (β , σ ) ∈ ak+1. ∗ 0 If σ ∈ ak, then the conclusion of (H) follows immediately from inductive hy- 0 ∗ pothesis (H) for ak. Otherwise we are in the case that max(ak) < σ and σ = σ. ∗ 0 0 Hence also β = β. Clearly min((ak ∪ bk) \ (τ + 1)) is equal to η. Since (β, η) ∈ bk 0 0 0 and max(ak) is the greatest member of ak ∪ bk less than η, inductive hypothesis (F) implies that β ≤ P ∩ κ, proving (H(i)). By inductive hypothesis (E),

Aη,P ∩κ = Aη∗,P ∩κ ∩ η = P ∩ η. Since θ ≤ η, Aη,P ∩κ ∩ θ = P ∩ θ.

As θ ∈ lim(Cη), Aθ,P ∩κ = Aη,P ∩κ ∩ θ. By Lemma 7.12(2), Aθ,P ∩κ = Aσ,P ∩κ ∩ θ. So P ∩ θ = Aη,P ∩κ ∩ θ = Aθ,P ∩κ = Aσ,P ∩κ ∩ θ. Since sup(N ∩ σ) = θ, it follows that

P ∩ sup(N ∩ σ) = Aσ,P ∩κ ∩ sup(N ∩ σ), which proves (H(ii)).

0 0 (I) Let (γ, ζ) ∈ ak+1 ∪ bk+1, where min(ak+1) = min(ak) < ζ. If (γ, ζ) ∈ ak, then the conclusion of (I) follows from inductive hypothesis (I). If (γ, ζ) ∈ ak+1 \ak, 0 then (γ, ζ) = (β, σ) and max(ak) < σ. Let P := Sk(Aη∗,β). Since (β, η) ∈ bk, by inductive hypothesis (I) we know that + P ∈ M ∩ Y,P ∩ κ = β, and P ∩ κ = Aη∗,β. 0 0 0 Also since max(ak) is the greatest member of ak∪bk less than η, inductive hypothesis (I) implies that 0 ∗ P ∩ N ∩ [max(ak), η ) 6= ∅. 0 0 0 But the greatest member of ak+1 ∪ bk+1 less than σ is also equal to max(ak), so we are done. 0 Suppose that (γ, ζ) ∈ bk and η < ζ. If ζ is not the second element of bk, then (I) follows immediately from inductive hypothesis (I). Suppose that ζ is the 0 0 0 second element of bk. Then η is the greatest member of ak ∪ bk less than ζ. Let P := Sk(Aη∗,γ ). By inductive hypothesis (I), + ∗ P ∈ M ∩ Y,P ∩ κ = γ, P ∩ κ = Aη∗,γ , and P ∩ N ∩ [η, η ) 6= ∅. − 0 0 Let ζ denote the largest member of ak+1 ∪ bk+1 less than ζ, and we will show that P ∩ N ∩ [ζ−, η∗) 6= ∅. If ζ− ≤ η, then this follows immediately from the fact that P ∩ N ∩ [η, η∗) 6= ∅. If A is nonempty, then clearly ζ− = max(A) < η, and we are done. MITCHELL’S THEOREM REVISITED 77

+ 0 − 0 Suppose that A is empty. If min((N ∩κ )\θ) ≤ max(ak), then ζ = max(ak) < 0 + η, and we are done. Suppose that max(ak) < σ = min((N ∩ κ ) \ θ). By inductive hypothesis (B), N ∩[η, ζ) 6= ∅. Since θ ≤ η, N ∩[θ, ζ) 6= ∅. As σ = min((N ∩κ+)\θ), it follows that σ < ζ, and so clearly ζ− = σ. If σ ≤ η, then we are done. Otherwise σ = min((N ∩ κ+) \ η). We know that P ∩ N ∩ [η, η∗) 6= ∅. But the first member of this intersection must be greater than or equal to min((N ∩ κ+) \ η) = σ. Hence P ∩ N ∩ [σ, η∗) 6= ∅, and we are done since σ = ζ−. In the final case, assume that (γ, ζ) is equal to (βδ, δ), for some δ ∈ A. By the definition of βδ, there exists Q ∈ M ∩ Y and ξ such that

Q ∩ κ = βδ, ξ ∈ Q ∩ N ∩ [σ, η), and δ = min(Cη \ (ξ + 1)). ∗ ∗ ∗ ∗ ∗ So Q ∩ η = Aη ,βδ . Since Q and η are closed under H , it follows that Aη ,βδ is closed under H∗.

∗ ∗ We will show that P := Sk(Aη ,βδ ) satisﬁes the conclusions of (I). Since Aη ,βδ is closed under H∗, + ∗ ∗ P ∩ κ = Aη ,βδ = Q ∩ η . Hence also

P ∩ κ = Q ∩ κ = βδ. Since η∗ is a limit point of Q, sup(P ) = sup(Q ∩ η∗) = η∗.

∗ As η and βδ are in M, so is P . ∗ To show that P ∈ Y, it suﬃces to show that lim(Cη∗ ) ∩ P is coﬁnal in η . Since Q ∩ η∗ = P ∩ η∗, η∗ is a limit point of Q. If η∗ is not in Q, then η∗ ∈ cl(Q) \ Q. By Lemma 7.13,

lim(Cη∗ ) ∩ Q = lim(Cη∗ ) ∩ P is cofinal in η∗. Otherwise η∗ ∈ Q. Since η∗ is a limit point of Q, cf(η∗) < κ. Therefore ot(Cη∗ ) ∈ Q ∩ κ by elementarity. Hence Cη∗ ⊆ Q by elementarity. Since ∗ ∗ η has uncountable cofinality, lim(Cη∗ ) is cofinal in η . So

lim(Cη∗ ) ∩ P = lim(Cη∗ ) ∩ Q = lim(Cη∗ ) is coﬁnal in η∗. − 0 0 Let δ denote the largest member of ak+1 ∪ bk+1 which is less than δ. Then either δ− = σ if δ = min(A), or else δ− is the largest member of A which is less than δ. In the ﬁrst case, the ordinal ξ, which is in Q ∩ N ∩ [σ, η), is a witness to the fact that Q ∩ N ∩ [δ−, η∗) 6= ∅. In the second case, ξ + 1 must be greater than − − δ , since otherwise δ = min(Cη \ (ξ + 1)) ≤ δ . So ξ + 1 is a witness to the fact that Q ∩ N ∩ [δ−, η∗) 6= ∅. In either case, since Q ∩ η∗ ⊆ P , P ∩ N ∩ [δ−, η∗) 6= ∅.

§13. Amalgamation of side conditions

We are now in a position to prove amalgamation results for S~-obedient side conditions over simple models in X , strong models in Y, and transitive models. The proofs of these results will use almost the entirety of the technology developed in the paper thus far. In Part III, the amalgamation results we present here will be used to prove the existence of strongly generic conditions. 78 THOMAS GILTON AND JOHN KRUEGER

Proposition 13.1. Let (A, B) be an S~-obedient side condition, where A ⊆ X and B ⊆ Y. Suppose that N ∈ A is simple, and (A, B) is closed under canonical models with respect to N. Assume that for all M ∈ A, if M < N then M ∩ N ∈ A. Let (C,D) be an S~-obedient side condition, where C ⊆ X and D ⊆ Y, such that A ∩ N ⊆ C ⊆ N and B ∩ N ⊆ D ⊆ N. Also assume that N 0 ∈ C is simple, and for all M ∈ A, if M < N, then there is M 0 in C such that M 0 < N 0,M ∩ N = M 0 ∩ N 0, and p(M,N) = p(M 0,N 0). Then (A ∪ C,B ∪ D) is an S~-obedient side condition. Proof. First note that for all M ∈ A, if M < N then M ∩ N ∈ C. For since N is simple, M ∩ N ∈ N by Lemma 8.2. So M ∩ N ∈ A ∩ N ⊆ C. Consider M < N in A. Since M ∩ N = M 0 ∩ N 0 and M 0 < N 0, it follows by Lemma 1.19(2) that 0 0 0 M ∩ βM,N = M ∩ N ∩ κ = M ∩ N ∩ κ = M ∩ βM 0,N 0 . 0 So M ∩ βM,N = M ∩ βM 0,N 0 . Also by Lemma 1.19(3), 0 0 βM,N = min(Λ \ sup(M ∩ N ∩ κ)) = min(Λ \ sup(M ∩ N ∩ κ)) = βM 0,N 0 .

So βM,N = βM 0,N 0 . To show that (A ∪ C,B ∪ D) is S~-obedient, we verify properties (1), (2), and (3) of Deﬁnition 5.3. (2) is immediate.

(3) Let M ∈ C and P ∈ B. Let β := P ∩κ, and suppose that ζ = min((M∩κ)\β). + Fix τ ∈ M ∩ P ∩ κ , and we will show that ζ ∈ Sτ . If β = ζ, then ζ ∈ Sτ since P is S~-strong. So assume that β < ζ, which means that β∈ / M. If P ∈ N then P ∈ B ∩ N ⊆ D, so ζ ∈ Sτ since (C,D) is S~-obedient. Assume that P/∈ N. Since M ∈ C and C ⊆ N, M ∈ N. Therefore ζ and τ are in N. Hence P ∩ κ = β < ζ < sup(N ∩ κ). Let ξ := min((N ∩ κ) \ β). Since M ⊆ N, ζ = min((M ∩ κ) \ ξ). By Lemma 10.9(1), there is Q ∈ B ∩ N ⊆ D such that Q ∩ κ = ξ and τ ∈ Q. Then ζ = min((M ∩ κ) \ (Q ∩ κ)) and τ ∈ M ∩ Q ∩ κ+.

So ζ ∈ Sτ since (C,D) is S~-obedient.

Let M ∈ A and P ∈ D. Let β := P ∩ κ, and suppose that ζ = min((M ∩ κ) \ β). + Fix τ ∈ M ∩ P ∩ κ , and we will show that ζ ∈ Sτ . If β = ζ, then ζ ∈ Sτ since P is S~-strong. So assume that β < ζ, which means that β∈ / M. Since P ∈ D and D ⊆ N, P ∈ N. So β = P ∩ κ is in N by elementarity.

Case 1: βM,N ≤ β. Since β ∈ N, ζ = min((M ∩ κ) \ β) is in RN (M). As P ∈ N ∩ Y is S~-strong, τ ∈ M ∩ P ∩ κ+, and sup(M ∩ ζ) < P ∩ κ = β < ζ, it follows that ζ ∈ Sτ since A is S~-adequate.

Case 2: β < βM,N ≤ ζ. Then ζ = min((M∩κ)\βM,N ). Since β ∈ (N∩βM,N )\M, it follows that M < N. Therefore ζ ∈ RN (M). As P ∈ N ∩ Y is S~-strong, τ ∈ M ∩ P ∩ κ+, and sup(M ∩ ζ) < P ∩ κ = β < ζ, MITCHELL’S THEOREM REVISITED 79 we have that ζ ∈ Sτ since A is S~-adequate.

Case 3: β < ζ < βM,N . Since β ∈ (N ∩ βM,N ) \ M, it follows that M < N. As M < N, P ∈ N ∩ Y, and

P ∩ κ < ζ < sup(M ∩ βM,N ) = sup(M ∩ N ∩ κ), it follows that M ∩ P ∩ κ+ ⊆ N by Lemma 8.7. In particular, τ ∈ M ∩ N ∩ κ+. As ζ < βM,N and M ∩ βM,N = M ∩ N ∩ κ, ζ = min((M ∩ N ∩ κ) \ β). But M ∩ N ∈ C, + P ∈ D, and τ ∈ (M ∩ N) ∩ P ∩ κ . So ζ ∈ Sτ since (C,D) is S~-obedient.

(1) Now we prove that A ∪ C is S~-adequate. By Proposition 1.29, A ∪ C is adequate. Let M ∈ A and L ∈ C. Then L ∈ N. We will prove that the remainder points in RM (L) and RL(M) are as required.

First, consider ζ ∈ RL(M). Then by Lemmas 2.4 and 2.5, either (1) M < N, ζ < βM,N , and ζ ∈ RL(M ∩ N), or (2) βM,N ≤ ζ and ζ ∈ RN (M).

Case 1: M < N, ζ < βM,N , and ζ ∈ RL(M ∩ N). Recall that L and M ∩ N are + in C. Fix τ ∈ L ∩ M ∩ κ , and we will show that ζ ∈ Sτ . Since L ∈ N, τ ∈ N. So τ ∈ L ∩ (M ∩ N). Since ζ ∈ RL(M ∩ N), it follows that ζ ∈ Sτ since C is S~-adequate. Suppose that P ∈ L ∩ Y is S~-strong, sup(M ∩ ζ) < P ∩ κ < ζ, and τ ∈ M ∩ P ∩ κ+.

We will show that ζ ∈ Sτ . Since P ∈ L and L ∈ N, P ∈ N. So P ∈ N ∩Y, M < N, and

P ∩ κ < ζ < sup(M ∩ βM,N ) = sup(M ∩ N ∩ κ). By Lemma 8.7, M ∩ P ∩ κ+ ⊆ N. In particular, τ ∈ N. So τ ∈ (M ∩ N) ∩ P ∩ κ+. Since ζ < βM,N and M < N, we have that M ∩ ζ = M ∩ N ∩ ζ. Therefore sup((M ∩ N) ∩ ζ) = sup(M ∩ ζ) < P ∩ κ < ζ.

So ζ ∈ RL(M ∩ N), P ∈ L ∩ Y is S~-strong, sup((M ∩ N) ∩ ζ) < P ∩ κ < ζ, and + τ ∈ (M ∩ N) ∩ P ∩ κ . It follows that ζ ∈ Sτ since C is S~-adequate.

+ Case 2: βM,N ≤ ζ and ζ ∈ RN (M). Fix τ ∈ L ∩ M ∩ κ , and we will show that + ζ ∈ Sτ . Since L ∈ N, τ ∈ N. So τ ∈ M ∩ N ∩ κ . As ζ ∈ RN (M), it follows that ζ ∈ Sτ since A is S~-adequate. Suppose that P ∈ L ∩ Y is S~-strong and sup(M ∩ ζ) < P ∩ κ < ζ. Fix τ ∈ + M ∩ P ∩ κ , and we will show that ζ ∈ Sτ . Since P ∈ L and L ∈ N, P ∈ N. So P ∈ N ∩Y is S~-strong. Since ζ ∈ RN (M), sup(M ∩ζ) < P ∩κ < ζ, and τ ∈ M ∩P , it follows that ζ ∈ Sτ since A is S~-adequate.

This completes the proof that the ordinals in RL(M) are as required.

Now consider ζ ∈ RM (L). Then by Lemmas 2.4 and 2.5, either M < N and ζ ∈ RM∩N (L), or there is ξ ∈ RM (N) such that ζ = min((L ∩ κ) \ ξ). Since ζ ∈ L and L ∈ N, ζ ∈ N. 80 THOMAS GILTON AND JOHN KRUEGER

+ Let τ ∈ L ∩ M ∩ κ , and we will show that ζ ∈ Sτ . Since L ∈ N, τ ∈ N. So τ ∈ L ∩ (M ∩ N). First, assume that M < N and ζ ∈ RM∩N (L). Since ζ ∈ RM∩N (L) and τ ∈ L ∩ (M ∩ N), it follows that ζ ∈ Sτ since C is S~-adequate. Secondly, assume that there is ξ ∈ RM (N) such that ζ = min((L ∩ κ) \ ξ). Since + ξ ∈ RM (N) and τ ∈ M ∩ N ∩ κ , by Lemma 10.9(2) there is Q ∈ B ∩ N ⊆ D such that Q ∩ κ = ξ and τ ∈ Q. Then ζ = min((L ∩ κ) \ (Q ∩ κ)) and τ ∈ L ∩ Q ∩ κ+, which implies that ζ ∈ Sτ since (C,D) is S~-obedient.

Suppose that P ∈ M ∩ Y is S~-strong, sup(L ∩ ζ) < P ∩ κ < ζ, and τ ∈ L ∩ P ∩ κ+.

We will prove that ζ ∈ Sτ . Note that since τ ∈ L and L ∈ N, τ ∈ N.

Case 1: βM,N ≤ P ∩ κ. Let θ := min((N ∩ κ) \ (P ∩ κ)). Note that θ exists since ζ ∈ N. Also ζ = min((L ∩ κ) \ θ). Since P ∩ κ ∈ (M ∩ κ) \ βM,N , it follows that θ ∈ RM (N) and sup(N ∩ θ) < P ∩ κ < θ. Also τ ∈ N ∩ P ∩ κ+. By Lemma 10.9(4), there is Q ∈ B ∩ N ⊆ D such that Q ∩ κ = θ and τ ∈ Q. But then ζ = min((L ∩ κ) \ (Q ∩ κ)) and τ ∈ L ∩ Q ∩ κ+, which implies that ζ ∈ Sτ since (C,D) is S~-obedient.

Case 2: P ∩ κ < βM,N and N ≤ M. Recall that either M < N and ζ ∈ RM∩N (L), or there is ξ ∈ RM (N) such that ζ = min((L ∩ κ) \ ξ). Since N ≤ M, we are in the second case.

Subcase 2(a): P ∩ κ < sup(N ∩ βM,N ). Since N ≤ M, sup(N ∩ βM,N ) = sup(M ∩ N ∩ κ). Therefore P ∩ κ < sup(M ∩ N ∩ κ). Since τ ∈ N ∩ P ∩ κ+, it + follows that τ ∈ M by Lemma 8.7. So τ ∈ M ∩ N ∩ κ . Since ξ ∈ RM (N), by Lemma 10.9(2) there is Q ∈ B ∩ N ⊆ D such that Q ∩ κ = ξ and τ ∈ Q. Since + ζ = min((L ∩ κ) \ (Q ∩ κ)) and τ ∈ L ∩ Q ∩ κ , it follows that ζ ∈ Sτ since (C,D) is S~-obedient.

Subcase 2(b): sup(N ∩ βM,N ) ≤ P ∩ κ. Since sup(N ∩ βM,N ) has countable coﬁnality and P ∩κ has uncountable coﬁnality, we have that sup(N ∩βM,N ) < P ∩κ. Let δ := min((N ∩ κ) \ βM,N ), which exists since ζ ∈ N. As N ≤ M, δ ∈ RM (N). Also sup(N ∩ δ) = sup(N ∩ βM,N ) < P ∩ κ < δ and τ ∈ N ∩P ∩κ+. By Lemma 10.9(4), there is Q ∈ B∩N ⊆ D such that Q∩κ = δ and τ ∈ Q. Since sup(N ∩ δ) < P ∩ κ < δ, we have that δ = min((N ∩ κ) \ (P ∩ κ)). As L ⊆ N and sup(L ∩ ζ) < P ∩ κ < ζ, clearly ζ = min((L ∩ κ) \ δ). So ζ = + min((L ∩ κ) \ (Q ∩ κ)) and τ ∈ L ∩ Q ∩ κ . It follows that ζ ∈ Sτ since (C,D) is S~-obedient.

Case 3: P ∩ κ < βM,N and M < N. Then P ∩ κ ∈ M ∩ βM,N = M ∩ N ∩ κ.

Subcase 3(a): τ < αM,N and P ∩ αM,N is bounded below αM,N . Then τ ∈ 0 P ∩ N ∩ αM,N . By Lemma 11.7, there is P ∈ M ∩ N ∩ Y which is S~-strong such that P 0 ∩ κ = P ∩ κ and τ ∈ P 0. Recall that either ζ ∈ RM∩N (L), or there is ξ ∈ RM (N) such that ζ = min((L ∩ 0 κ) \ ξ). Suppose ﬁrst that ζ ∈ RM∩N (L). Since P ∈ M ∩ N ∩ Y is S~-strong, MITCHELL’S THEOREM REVISITED 81

0 0 + sup(L ∩ ζ) < P ∩ κ = P ∩ κ < ζ, and τ ∈ L ∩ P ∩ κ , it follows that ζ ∈ Sτ since C is S~-adequate. Now suppose that there is ξ ∈ RM (N) such that ζ = min((L ∩ κ) \ ξ). Then by Lemma 10.9(3), there is Q ∈ B ∩ N ⊆ D such that Q ∩ κ = ξ and N ∩ P 0 ∩ κ+ ⊆ Q. + So ζ = min((L ∩ κ) \ (Q ∩ κ)) and τ ∈ L ∩ Q ∩ κ . It follows that ζ ∈ Sτ since (C,D) is S~-obedient.

Case 3(b): Either τ < αM,N and P ∩αM,N is unbounded in αM,N , or αM,N ≤ τ. In the ﬁrst case, we apply Lemma 11.6 letting τ0 = τ to get that there exists Q ∈ N 0 ∩ Y which is S~-strong such that Q ∩ κ = P ∩ κ and τ ∈ Q. In the second case, we apply the main proxy lemma, Lemma 11.5. Since τ ∈ P and P ∈ M,

αM,N ≤ τ < sup(P ) < sup(M). + + Let η := min((M ∩ κ ) \ τ), which is in RN (M). Note that the assumptions of Lemma 11.5 for η∗ = η are satisﬁed. By Lemma 11.5(1), there is Q ∈ N 0 ∩ Y which is S~-strong such that Q ∩ κ = P ∩ κ and τ ∈ Q. In either case, we have that Q ∈ N 0 ∩ Y is S~-strong, Q ∩ κ = P ∩ κ, and τ ∈ Q. 0 Let us note that if ζ ∈ RN 0 (L), then ζ ∈ Sτ and we are done. For Q ∈ N ∩ Y is S~-strong, sup(L ∩ ζ) < P ∩ κ = Q ∩ κ < ζ, + and τ ∈ L ∩ Q ∩ κ . It follows that ζ ∈ Sτ since C is S~-adequate.

Subcase 3(b(i)): βL,N 0 ≤ ζ. We claim that ζ ∈ RN 0 (L), which ﬁnishes the 0 proof. If βL,N 0 ≤ P ∩ κ, then P ∩ κ = Q ∩ κ ∈ (N ∩ κ) \ βL,N 0 . Therefore ζ = min((L ∩ κ) \ (P ∩ κ)) is in RN 0 (L). Suppose on the other hand that P ∩ κ < βL,N 0 . 0 0 Then ζ = min((L∩κ)\βL,N 0 ). Since P ∩κ ∈ (N ∩βL,N 0 )\L, we have that L < N . So ζ ∈ RN 0 (L).

Subcase 3(b(ii)): ζ < βL,N 0 . In particular, since Q ∩ κ = P ∩ κ < ζ and ζ ∈ L ∩ βL,N 0 , it follows that

Q ∩ κ < sup(L ∩ βL,N 0 ) < βL,N 0 . 0 0 0 0 As Q ∩ κ ∈ (N ∩ βL,N 0 ) \ L, we have that L < N . So L < N , Q ∈ N ∩ Y, and 0 Q ∩ κ < sup(L ∩ βL,N 0 ) = sup(L ∩ N ∩ κ). By Lemma 8.7, Q ∩ L ∩ κ+ ⊆ N 0. Since τ ∈ L ∩ Q ∩ κ+, it follows that τ ∈ N 0. By Lemma 11.5(2) in the case that αM,N ≤ τ, and by Lemma 11.6(2) in the case that 0 0 0 τ < αM,N , there is P ∈ M ∩ Y which is S~-strong such that P ∩ κ = P ∩ κ and τ ∈ P 0. By Lemmas 1.27(1) and 2.6(1), either βL,M = βL,M∩N = βL,M 0 , or βM,N < βL,M 0 . Suppose ﬁrst that βL,M = βL,M 0 . We claim that

βL,M 0 ≤ P ∩ κ.

Suppose for a contradiction that P ∩ κ < βL,M 0 = βL,M . Since Λ is coﬁnal in P ∩ κ by the elementarity of P , and sup(L ∩ ζ) < P ∩ κ, we can ﬁnd π ∈ Λ ∩ P ∩ κ such that sup(L ∩ ζ) < π. Then π < βL,M . By Lemma 1.19(5),

L ∩ M ∩ [π, βL,M ) 6= ∅. 82 THOMAS GILTON AND JOHN KRUEGER

Fix ξ in this intersection. As ζ ∈ RM (L), βL,M ≤ ζ. So ξ ∈ L and

sup(L ∩ ζ) < π ≤ ξ < βL,M ≤ ζ.

But sup(L ∩ ζ) < ξ < ζ and ξ ∈ L is obviously impossible. Hence indeed βL,M 0 ≤ P ∩ κ. 0 0 0 So P ∩ κ = P ∩ κ ∈ (M ∩ κ) \ βL,M 0 . Since ζ = min((L ∩ κ) \ (P ∩ κ)), we 0 0 0 have that ζ ∈ RM 0 (L). As P ∈ M ∩ Y is S~-strong, sup(L ∩ ζ) < P ∩ κ < ζ, and 0 + τ ∈ L ∩ P ∩ κ , it follows that ζ ∈ Sτ since C is S~-adequate. The other alternative is that βM,N < βL,M 0 . We will show that this is impossible. 0 0 0 So assume that βM,N < βL,M 0 . We claim that L < M . For P ∩ κ = P ∩ κ ∈ M . Also

P ∩ κ < βM,N < βL,M 0 . So 0 P ∩ κ ∈ (M ∩ βL,M 0 ) \ L, which implies that L < M 0. Next we claim that βM,N ≤ ζ. Suppose for a contradiction that ζ < βM,N . Then 0 0 ζ ∈ L ∩ βM,N ⊆ L ∩ βL,M 0 , and L ∩ βL,M 0 ⊆ M since L < M . So 0 0 ζ ∈ M ∩ βM,N = M ∩ βM 0,N 0 = M ∩ βM,N .

Hence ζ ∈ M. But this is not true since ζ ∈ RM (L). Since P ∩ κ < βM,N ≤ ζ and sup(L ∩ ζ) < P ∩ κ, clearly ζ = min((L ∩ κ) \ βM,N ). Since βM,N < βL,M 0 and βM,N ∈ Λ, by Lemma 1.19(5) we have that 0 L ∩ M ∩ [βM,N , βL,M 0 ) is nonempty. Since ζ = min((L ∩ κ) \ βM,N ), it follows that 0 0 ζ < βL,M 0 . As L < M , ζ ∈ M . Now we will get a contradiction. By Subcase 3(b(ii)), ζ < βL,N 0 . So ζ ∈ L∩βL,N 0 . Since L < M 0 < N 0, we have that L < N 0. Hence ζ ∈ N 0. So ζ ∈ M 0 ∩N 0 ∩κ, which implies that ζ < βM 0,N 0 = βM,N . But βM,N ≤ ζ, and we have a contradiction. Proposition 13.2. Let (A, B) be an S~-obedient side condition, where A ⊆ X and B ⊆ Y. Suppose that P ∈ B satisﬁes that cf(sup(P )) = P ∩ κ. Assume that for all M ∈ A, M ∩ P ∈ A, and for all Q ∈ B, if Q ∩ κ < P ∩ κ then Q ∩ P ∈ B. Let (C,D) be an S~-obedient side condition, where C ⊆ X and D ⊆ Y, such that A ∩ P ⊆ C ⊆ P and B ∩ P ⊆ D ⊆ P. In addition, assume that there exists P 0 ∈ D such that cf(sup(P 0)) = P 0 ∩ κ, and for all M ∈ A, there exists M 0 ∈ C such that M ∩ P = M 0 ∩ P 0 and p(M,P ) = p(M 0,P 0). Then (A ∪ C,B ∪ D) is an S~-obedient side condition. Proof. By Lemma 8.3, P and P 0 are simple. Note that for all M ∈ A, M ∩ P ∈ C. For M ∩ P ∈ P by Lemma 8.4, and so M ∩ P ∈ A ∩ P ⊆ C. Similarly, for all Q ∈ B with Q ∩ κ < P ∩ κ, Q ∩ P ∈ D. For Q ∩ P ∈ P by Lemma 8.4, and hence Q ∩ P ∈ B ∩ P ⊆ D. Let β := P ∩ κ and β0 := P 0 ∩ κ. Consider M ∈ A. Then by our assumptions, M ∩ β = M ∩ P ∩ κ = M 0 ∩ P 0 ∩ κ = M 0 ∩ β0. So M ∩β = M 0 ∩β0. In particular, since β0 has uncountable coﬁnality, sup(M ∩β) < β0. MITCHELL’S THEOREM REVISITED 83

To show that (A ∪ C,B ∪ D) is S~-obedient, we verify properties (1), (2), and (3) of Deﬁnition 5.3. (2) is immediate.

(3) Let M ∈ A and Q ∈ D. Let θ := Q∩κ, and suppose that ζ = min((M ∩κ)\θ). + Let τ ∈ Q ∩ M ∩ κ , and we will show that ζ ∈ Sτ .

Case 1: P ∩ κ ≤ ζ. Since Q ∈ P , θ < P ∩ κ, so ζ = min((M ∩ κ) \ (P ∩ κ)). As + τ ∈ Q and Q ∈ P , τ ∈ P . So τ ∈ P ∩ M ∩ κ . Therefore ζ ∈ Sτ since (A, B) is S~-obedient.

Case 2: ζ < P ∩ κ. Then ζ ∈ M ∩ P ∩ κ. Since M ∩ P ∩ κ = M ∩ β is an initial segment of M ∩ κ, ζ = min((M ∩ P ∩ κ) \ θ). As τ ∈ Q and Q ∈ P , τ ∈ P . + So τ ∈ Q ∩ (M ∩ P ) ∩ κ . Since M ∩ P ∈ C, it follows that ζ ∈ Sτ as (C,D) is S~-obedient.

Let M ∈ C and Q ∈ B. Let θ := Q ∩ κ, and suppose that ζ = min((M ∩ κ) \ θ). + Fix τ ∈ Q ∩ M ∩ κ , and we will show that ζ ∈ Sτ . Since ζ ∈ M and M ∈ P , ζ ∈ P ∩ κ. Hence Q ∩ κ < P ∩ κ. So by our assumptions, Q ∩ P ∈ D. As τ ∈ M and M ∈ P , τ ∈ P . So τ ∈ (Q ∩ P ) ∩ M ∩ κ+. Since Q ∩ P ∩ κ = Q ∩ κ = θ, ζ = min((M ∩ κ) \ (Q ∩ P ∩ κ)).

It follows that ζ ∈ Sτ since (C,D) is S~-obedient.

(1) The set A ∪ C is adequate by Proposition 1.35. Let M ∈ A and L ∈ C. Then L ∈ P . We will prove that the remainder points in RM (L) and RL(M) are as required.

Consider ζ ∈ RL(M). Then by Lemma 2.9, either ζ ∈ RL(M ∩ P ) or ζ = min((M ∩ κ) \ β).

+ Case 1: ζ ∈ RL(M ∩ P ). Fix τ ∈ L ∩ M ∩ κ , and we will show that ζ ∈ Sτ . + Since τ ∈ L and L ∈ P , τ ∈ P . So τ ∈ L ∩ (M ∩ P ) ∩ κ . Since ζ ∈ RL(M ∩ P ), it follows that ζ ∈ Sτ since C is S~-adequate. Suppose that Q ∈ L ∩ Y is S~-strong and sup(M ∩ ζ) < Q ∩ κ < ζ. + Fix τ ∈ Q ∩ M ∩ κ , and we will show that ζ ∈ Sτ . Since M ∩ P ∩ κ = M ∩ β is an initial segment of M ∩ κ and ζ ∈ M ∩ P ∩ κ, sup(M ∩ P ∩ ζ) = sup(M ∩ ζ) < Q ∩ κ < ζ. As Q ∈ L and L ∈ P , Q ∈ P . And since τ ∈ Q and Q ∈ P , τ ∈ P . So + τ ∈ Q ∩ (M ∩ P ) ∩ κ . Since ζ ∈ RL(M ∩ P ), it follows that ζ ∈ Sτ since C is S~-adequate.

Case 2: ζ = min((M ∩ κ) \ β). Fix τ ∈ L ∩ M ∩ κ+, and we will show that + ζ ∈ Sτ . Since τ ∈ L and L ∈ P , τ ∈ P . So τ ∈ P ∩ M ∩ κ . Since M ∈ A and P ∈ B, it follows that ζ ∈ Sτ since (A, B) is S~-obedient. Suppose that Q ∈ L ∩ Y is S~-strong and sup(M ∩ ζ) < Q ∩ κ < ζ. 84 THOMAS GILTON AND JOHN KRUEGER

+ Fix τ ∈ Q ∩ M ∩ κ , and we will show that ζ ∈ Sτ . As Q ∈ L and L ∈ P , Q ∈ P . And since τ ∈ Q and Q ∈ P , τ ∈ P . So τ ∈ M ∩ P ∩ κ+. As M ∈ A and P ∈ B, it follows that ζ ∈ Sτ since (A, B) is S~-obedient.

+ Consider ζ ∈ RM (L). By Lemma 2.9, ζ ∈ RM∩P (L). Fix τ ∈ L ∩ M ∩ κ , and + we will show that ζ ∈ Sτ . Since τ ∈ L and L ∈ P , τ ∈ P . So τ ∈ L ∩ (M ∩ P ) ∩ κ . As ζ ∈ RM∩P (L), it follows that ζ ∈ Sτ since C is S~-adequate. Suppose that Q ∈ M ∩ Y is S~-strong and sup(L ∩ ζ) < Q ∩ κ < ζ. + Fix τ ∈ Q∩L∩κ , and we will show that ζ ∈ Sτ . Since ζ ∈ L and L ∈ P , ζ ∈ P ∩κ. Therefore Q ∩ κ ∈ M ∩ P ∩ κ. As τ ∈ L and L ∈ P , τ ∈ P . So τ ∈ Q ∩ P ∩ κ+.

Case 1: τ < αM,P and Q∩αM,P is bounded below αM,P . Then τ ∈ Q∩P ∩αM,P . By Lemma 11.7, there is Q0 ∈ M ∩P ∩Y which is S~-strong such that Q0 ∩κ = Q∩κ and τ ∈ Q0. So sup(L ∩ ζ) < Q0 ∩ κ = Q ∩ κ < ζ 0 + 0 and τ ∈ Q ∩ L ∩ κ . Since ζ ∈ RM∩P (L) and Q ∈ (M ∩ P ) ∩ Y is S~-strong, it follows that ζ ∈ Sτ since C is S~-adequate.

Case 2: Either τ < αM,P and Q ∩ αM,P is unbounded in αM,P , or αM,P ≤ τ. In the ﬁrst case, we apply Lemma 11.6(1) to get Q∗ ∈ P 0 ∩ Y such that τ ∈ Q∗. In the second case, we apply the main proxy lemma, Lemma 11.5. Assuming αM,P ≤ τ, let η := min((M ∩ κ+) \ τ). Note that η exists since sup(Q) ∈ M and τ < sup(Q). Also τ ∈ Q ∩ P ∩ [sup(M ∩ η), η). By Lemma 11.5(1) there is Q∗ ∈ P 0 ∩ Y such that τ ∈ Q∗. Thus in either case, there is Q∗ ∈ P 0 ∩ Y such that τ ∈ Q∗. As τ ∈ Q∗ and Q∗ ∈ P 0, τ ∈ P 0. Also by Lemma 11.6(2) in the ﬁrst case, and Lemma 11.5(3) in the second case, there is Q0 ∈ M 0 ∩ Y such that Q0 is S~-strong, Q0 ∩ κ = Q ∩ κ, and τ ∈ Q0. 0 By Lemma 2.10(2), either βL,M = βL,M 0 or β < βL,M 0 . First, suppose that βL,M = βL,M 0 . Since ζ ∈ RM∩P (L), Lemma 2.10(3) implies that ζ ∈ RM 0 (L). But Q0 ∈ M 0 ∩ Y is S~-strong, sup(L ∩ ζ) < Q0 ∩ κ = Q ∩ κ < ζ, 0 + and τ ∈ Q ∩ L ∩ κ . It follows that ζ ∈ Sτ since C is S~-adequate. 0 Secondly, assume that β < βL,M 0 . Since Q ∩ κ ∈ M ∩ P ∩ κ = M ∩ β = M 0 ∩ β0, it follows that 0 Q ∩ κ < β < βL,M 0 . 0 0 0 As Q ∩ κ = Q ∩ κ ∈ (M ∩ βL,M 0 ) \ L, we have that L < M . We claim that β0 ≤ ζ. Otherwise since L < M 0, 0 0 ζ ∈ L ∩ β ⊆ L ∩ βL,M 0 ⊆ M . 0 0 So ζ ∈ M ∩ β = M ∩ β. Hence ζ ∈ M, which contradicts that ζ ∈ RM (L). Since sup(L ∩ ζ) < Q ∩ κ < β0 ≤ ζ, MITCHELL’S THEOREM REVISITED 85 we have that ζ = min((L ∩ κ) \ β0). As noted above, τ ∈ P 0. So τ ∈ L ∩ P 0 ∩ κ+, and ζ = min((L ∩ κ) \ β0) = min((L ∩ κ) \ (P 0 ∩ κ)). ~ It follows that ζ ∈ Sτ since (C,D) is S-obedient.

Proposition 13.3. Let (A, B) be an S~-obedient side condition, where A ⊆ X and B ⊆ Y. Suppose that X ≺ A is such that |X| = κ, X ∩ κ+ ∈ κ+, and X<κ ⊆ X. Let θ := X ∩ κ+. Let M0,...,Mk−1 and P0,...,Pm−1 enumerate the members of A and B re- i ~ spectively. For each i < k, let hQn : n < ωi enumerate the S-strong models in Mi ∩ Y. Let (C,D) be an S~-obedient side condition, where C ⊆ X and D ⊆ Y, such that A ∩ X ⊆ C ⊆ X and B ∩ X ⊆ D ⊆ X.

0 0 0 0 0 i Assume that M0,...,Mk−1, P0,...,Pm−1, θ , and hRn : n < ωi for i < k satisfy the following properties: (1) θ0 ∈ θ ∩ cof(κ); 0 0 0 (2) for all i < k, Mi ∈ C and Mi ∩ θ = Mi ∩ θ ; 0 0 0 (3) for all j < m, Pj ∈ D and Pj ∩ θ = Pj ∩ θ ; i ~ 0 (4) for all i < k, hRn : n < ωi enumerates the S-strong models in Mi ∩ Y, and i i 0 for all n < ω, Qn ∩ θ = Rn ∩ θ . Then (A ∪ C,B ∪ D) is an S~-obedient side condition.

Proof. To show that (A∪C,B ∪D) is S~-obedient, we verify properties (1), (2), and (3) of Deﬁnition 5.3. (2) is immediate.

0 0 (3) Let M ∈ C and P ∈ B. Fix j < m such that P = Pj, and let P := Pj. Let β := P ∩ κ, and suppose that ζ := min((M ∩ κ) \ β). Fix τ ∈ M ∩ P ∩ κ+, + and we will show that ζ ∈ Sτ . Since τ ∈ M and M ∈ X, τ ∈ X ∩ κ = θ. So τ ∈ P ∩ θ = P 0 ∩ θ0. Also P 0 ∩ κ = P ∩ κ = β. Hence τ ∈ M ∩ P 0 ∩ κ+ and 0 ζ = min((M ∩ κ) \ (P ∩ κ)). It follows that ζ ∈ Sτ since (C,D) is S~-obedient. 0 0 Let M ∈ A and P ∈ D. Fix i < k such that M = Mi, and let M := Mi . Let β := P ∩ κ, and suppose that ζ = min((M ∩ κ) \ β). Fix τ ∈ M ∩ P ∩ κ+, and + we will show that ζ ∈ Sτ . Since τ ∈ P and P ∈ X, τ ∈ X ∩ κ = θ. Hence τ ∈ M ∩ θ = M 0 ∩ θ0. Also M ∩ κ = M 0 ∩ κ, so ζ = min((M 0 ∩ κ) \ β). Since 0 + τ ∈ M ∩ P ∩ κ , it follows that ζ ∈ Sτ since (C,D) is S~-obedient.

(1) The set A ∪ C is adequate by Proposition 1.38. Let M ∈ A and L ∈ C. We will prove that the remainder points in RM (L) and RL(M) are as required. Fix 0 0 i < k such that M = Mi, and let M := Mi . 0 0 Consider ζ ∈ RL(M). Since M ∩ κ = M ∩ κ, ζ ∈ RL(M ) by Lemma 2.11. + Let τ ∈ L ∩ M ∩ κ , and we will show that ζ ∈ Sτ . Since τ ∈ L and L ∈ X, τ ∈ X ∩ κ+ = θ. Hence τ ∈ M ∩ θ = M 0 ∩ θ0. Therefore τ ∈ L ∩ M 0. Since 0 ζ ∈ RL(M ), it follows that ζ ∈ Sτ as C is S~-adequate. Suppose that P ∈ L ∩ Y is S~-strong and sup(M ∩ ζ) < P ∩ κ < ζ. 86 THOMAS GILTON AND JOHN KRUEGER

+ Let τ ∈ P ∩ M ∩ κ , and we will show that ζ ∈ Sτ . Since P ∈ L and L ∈ X, P ∈ X. And as τ ∈ P and P ∈ X, τ ∈ X ∩ κ+ = θ. Hence τ ∈ M ∩ θ = M 0 ∩ θ0. Therefore τ ∈ P ∩ M 0 ∩ κ+. Also since M ∩ κ = M 0 ∩ κ, sup(M 0 ∩ ζ) = sup(M ∩ ζ) < P ∩ κ < ζ.

0 As ζ ∈ RL(M ), it follows that ζ ∈ Sτ since C is S~-adequate.

0 Now consider ζ ∈ RM (L). Since M ∩ κ = M ∩ κ, ζ ∈ RM 0 (L) by Lemma 2.11. + Let τ ∈ L ∩ M ∩ κ , and we will show that ζ ∈ Sτ . Since τ ∈ L and L ∈ X, τ ∈ X ∩ κ+ = θ. Hence τ ∈ M ∩ θ = M 0 ∩ θ0. Therefore τ ∈ L ∩ M 0. Since ζ ∈ RM 0 (L), it follows that ζ ∈ Sτ since C is S~-adequate. Let P ∈ M ∩ Y be S~-strong, and assume that sup(L ∩ ζ) < P ∩ κ < ζ. + Let τ ∈ L ∩ P ∩ κ , and we will show that ζ ∈ Sτ . Since τ ∈ L and L ∈ X, + i τ ∈ X ∩ κ = θ. Fix n < ω such that P = Qn. Then i i 0 τ ∈ P ∩ θ = Qn ∩ θ = Rn ∩ θ . i + i 0 ~ So τ ∈ Rn ∩ L ∩ κ . Now Rn ∈ M ∩ Y is S-strong, and i i sup(L ∩ ζ) < P ∩ κ = Qn ∩ κ = Rn ∩ κ < ζ. ~ Since ζ ∈ RM 0 (L), it follows that ζ ∈ Sτ since C is S-adequate.

Part 3. Mitchell’s Theorem

§14. The ground model

With the general development of side conditions from Parts I and II at our disposal, we now begin our proof of Mitchell’s theorem. We start by describing the ground model over which we will force a generic extension satisfying that there is no stationary subset of ω2 ∩ cof(ω1) in the approachability ideal I[ω2].

We will use the same notation which was introduced at the beginning of Parts I and II, together with some additional assumptions. Recall that κ ≥ ω2 is regular, κ + + 2 = κ , and κ. Also the cardinal λ from Part I is equal to κ . In addition, we will assume that κ is a greatly Mahlo cardinal, and the thin stationary set T ∗ from

Notation 1.4 is equal to Pω1 (κ). + Deﬁne a sequence of sets hSξ : ξ < κ i inductively as follows. Let S0 denote the + set of inaccessible cardinals less than κ. Let δ < κ , and suppose that Sξ has been deﬁned for all ξ < δ. If δ = δ0 + 1, then let α ∈ Sδ if α is inaccessible, α ∈ Sδ0 , and

Sδ0 ∩ α is stationary in α. If δ is a limit ordinal, then let α ∈ Sδ if α is inaccessible, + and for all ξ ∈ Aδ,α, α ∈ Sξ. Let S~ := hSξ : ξ < κ i. + The fact that κ is greatly Mahlo implies that for all δ < κ , Sδ is stationary in κ. In fact, it is easily seen that this consequence is actually equivalent to κ being greatly Mahlo. See [1, Deﬁnition 4.2] for more information about greatly Mahlo cardinals. MITCHELL’S THEOREM REVISITED 87

Notation 14.1. For the remainder of Part III, the structure A from Notation 7.6 will be equal to + ∗ ∗ ∗ ∗ ~ ~ ∗ ~ (H(κ ), ∈, E, κ, T , π ,C , Λ, Y0, f , C, A, c , S).

In Proposition 7.20, we proved that the set of simple models in X is stationary in + Pω1 (H(κ )). We will prove in Proposition 15.3 that most simple models in X have strongly generic conditions. The next proposition describes the kind of models in Y which will have strongly generic conditions.

+ Proposition 14.2. There are stationarily many P ∈ Pκ(H(κ )) such that P ∈ Y, P is S~-strong, and cf(sup(P )) = P ∩ κ. Proof. Let F : H(κ+)<ω → H(κ+). Fix X which is an elementary substructure of A of size κ such that X is closed under F , and τ := X ∩ κ+ has cofinality κ. Note that X = Sk(X ∩ κ+) = Sk(τ). Since τ is the union of the increasing and continuous sequence of sets {Aτ,i : i < κ}, it follows that X is the union of the increasing and continuous sequence of sets {Sk(Aτ,i): i < κ}. For all infinite β < κ, |Aτ,β| ≤ |β| < κ by Notation 7.4(3). Since τ has cofinality κ, sup(Aτ,β) < τ, and hence sup(Aτ,β) ∈ X. Fix a club C ⊆ κ such that for all ∗ α ∈ C, Aτ,α is closed under H , Aτ,α ∩ κ = α, Sk(Aτ,α) is closed under F , and for all β < α, sup(Aτ,β) ∈ Aτ,α. As Sτ is stationary in κ, we can fix α ∈ lim(C) ∩ Sτ . Let P := Sk(Aτ,α). We claim that P ∈ Y, P is S~-strong, cf(sup(P )) = P ∩ κ, and P is closed under F . The last statement follows from the fact that α ∈ C. Since Aτ,α is closed under ∗ + H , P ∩ κ = Aτ,α. In particular, since α ∈ C, P ∩ κ = α. As α ∈ Sτ , α is inaccessible. After we show that cf(sup(P )) = α, it will follow that P ∈ Y by Lemma 7.15. + + To show that P is S~-strong, let σ ∈ P ∩ κ . Then σ ∈ P ∩ κ = Aτ,α. Since α ∈ Sτ and τ is a limit ordinal, for all π ∈ Aτ,α, α ∈ Sπ. In particular, α ∈ Sσ. It remains to show that cf(sup(P )) = α. For α0 < α1 in C ∩ α,

sup(Aτ,α0 ) ∈ Aτ,α1 ⊆ P by the deﬁnition of C. Since P ∩ κ is a limit point of C, [ Aτ,α = {Aτ,β : β ∈ C ∩ α}. Therefore sup(P ) = sup(Aτ,α) = sup{sup(Aτ,β): β ∈ C ∩ α}, which is the supremum of a strictly increasing sequence. Since α is in Sτ , α is inaccessible, so C ∩ α has order type α. Hence sup(P ) has coﬁnality equal to P ∩ κ = α.

§15. The forcing poset

We now deﬁne and analyze the forcing poset which will force that there is no stationary subset of ω2 ∩ cof(ω1) in the approachability ideal I[ω2].

Deﬁnition 15.1. Let P be the forcing poset consisting of pairs p = (Ap,Bp) satisfying:

(1) Ap ⊆ X , and for all M ∈ Ap, M ≺ (A, Y); 88 THOMAS GILTON AND JOHN KRUEGER

(2) Bp ⊆ Y; (3) (Ap,Bp) is an S~-obedient side condition.

Let q ≤ p if Ap ⊆ Aq and Bp ⊆ Bq. The rest of this section is devoted to proving amalgamation results for P, which in turn yield the existence of strongly generic conditions.

Lemma 15.2. Let N ∈ X with N ≺ (A, Y). Then qN := ({N}, ∅) is in P, and for all p ∈ N ∩ P, p and qN are compatible. Proof. Immediate from Lemma 5.4(1).

Proposition 15.3. Let N ∈ X be simple such that N ≺ (A, Y, P). Let qN := ({N}, ∅). Then qN is a universal strongly N-generic condition. See Section 3 for a discussion of universal strongly generic conditions.

Proof. By Lemma 15.2, qN is compatible with all conditions in N ∩P. So it suﬃces to show that qN is strongly N-generic. Let r0 ≤ qN be given. We will ﬁnd a condition v in N ∩ P such that for all w ≤ v in N ∩ P, r0 and w are compatible.

Let M0,...,Mk−1 list the models M in Ar0 \ N such that M < N. Note that by Lemma 8.2, Mi ∩ N ∈ N for all i < k. By ﬁnitely many applications of Lemmas 5.5(1) and 7.16, together with the fact that N ≺ (A, Y), the pair

r1 := (Ar0 ∪ {M0 ∩ N,...,Mk−1 ∩ N},Br0 ) is a condition below r0.

By Proposition 10.8, there is a condition r ≤ r1 such that Ar = Ar1 , and (Ar,Br) is closed under canonical models with respect to N. Note that the assumptions of the ﬁrst paragraph of Proposition 13.1 hold for A = Ar and B = Br. The objects r, N, and M0,...,Mk−1 witness that the following statement holds in (A, Y, P): 0 0 0 There exist v, N , and M0,...,Mk−1 satisfying: (1) v ∈ P; 0 0 0 (2) Ar ∩ N ⊆ Av, Br ∩ N ⊆ Bv, and M0,...,Mk−1 and N are in Av; (3) N 0 is simple; 0 0 0 0 0 0 (4) for all i < k, Mi < N , Mi ∩ N = Mi ∩ N , and p(Mi,N) = p(Mi ,N ).

The parameters which appear in the above statement, namely Ar ∩ N, Br ∩ N, and for i < k, Mi ∩ N and p(Mi,N), are all members of N. By the elementarity of 0 0 0 N, there are v, N , and M0,...,Mk−1 in N which satisfy the same statement.

We will show that for all w ≤ v in N ∩ P, w is compatible with r, and hence is compatible with r0 since r ≤ r0. This will complete the proof. So ﬁx w ≤ v in N ∩ P. We claim that the pair

(Ar ∪ Aw,Br ∪ Bw) is in P. Note that the assumptions of the second paragraph of Proposition 13.1 hold for C = Aw and D = Bw. So by Proposition 13.1, (Ar ∪ Aw,Br ∪ Bw) is an S~-obedient side condition. So this pair is a condition in P, and it is obviously below r and w. MITCHELL’S THEOREM REVISITED 89

Corollary 15.4. The forcing poset P satisﬁes the ω1-covering property. In particular, it preserves ω1. + Proof. By Proposition 7.20, the set of N ∈ Pω1 (H(κ )) such that N ∈ X and N is simple is stationary. Hence there are stationarily many such N with N ≺ (A, Y, P). Any such N has a universal strongly N-generic condition by Proposition 15.3. By Corollary 3.10, P has the ω1-covering property.

Next we prove that many models in Y have strongly generic conditions. ~ Lemma 15.5. Let P ∈ Y be S-strong. Let qP := (∅, {P }). Then qP is in P, and for all p ∈ P ∩ P, p and qP are compatible. Proof. Immediate from Lemma 5.4(2). Proposition 15.6. Let P ∈ Y be S~-strong such that cf(sup(P )) = P ∩ κ and P ≺ (A, Y, P). Let qP := (∅, {P }). Then qP is a universal strongly P -generic condition.

Proof. By Lemma 15.5, qP is compatible with all members of P ∩P. So it suffices to show that qP is strongly P -generic. Let r0 ≤ qP be given. We will find a condition v ∈ P ∩ P such that for all w ≤ v in P ∩ P, r0 and w are compatible. By finitely many applications of Lemmas 5.5(2), 5.5(3), and 7.16, together with the fact that P ≺ (A, Y), there is a condition r ≤ r0 such that

Ar = Ar0 ∪ {P ∩ M : M ∈ Ar0 }, and

Br = Br0 ∪ {P ∩ Q : Q ∈ Br0 ,Q ∩ κ < P ∩ κ}. Then the assumptions of the ﬁrst paragraph of Proposition 13.2 hold for A = Ar and B = Br. Let β := P ∩ κ. Let M0,...,Mk−1 list the members of Ar. The objects r, P , β, and M0,...,Mk−1 witness that the following statement holds in (A, Y, P): 0 0 0 0 There exist v, P , β , and M0,...,Mk−1 satisfying: (1) v ∈ P; 0 0 0 (2) Ar ∩ P ⊆ Av, Br ∩ P ⊆ Bv, M0,...,Mk−1 are in Av, and P ∈ Bv; (3) P 0 ∩ κ = β0 and cf(sup(P 0)) = β0; 0 0 0 0 (4) for all i < k, Mi ∩ P = Mi ∩ P and p(Mi,P ) = p(Mi ,P ).

The parameters appearing in the statement above, namely, Ar ∩ P , Br ∩ P , and for i < k, Mi ∩ P and p(Mi,P ), are all members of P . By the elementarity of P , 0 0 0 0 we can ﬁx v, P , β , and M0,...,Mk−1 in P which satisfy the same statement. 0 0 For each M ∈ Ar, let M denote Mi , where i < k and M = Mi.

We will show that for all w ≤ v in P ∩ P, w is compatible with r, and hence is compatible with r0 since r ≤ r0. This will complete the proof. So ﬁx w ≤ v in P ∩ P. We claim that the pair

(Ar ∪ Aw,Br ∪ Bw) is in P. Note that the assumptions of the second paragraph of Proposition 13.2 hold for C = Aw and D = Bw. So by Proposition 13.2, (Ar ∪ Aw,Br ∪ Bw) is an 90 THOMAS GILTON AND JOHN KRUEGER

S~-obedient side condition. So this pair is a condition in P, and it is obviously below r and w. Corollary 15.7. The forcing poset P has the κ-covering property. In particular, P forces that κ is a regular cardinal. Proof. By Proposition 14.2, there are stationarily many P in Y such that P is S~- strong and cf(sup(P )) = P ∩ κ. Therefore there are stationarily many such P with P ≺ (A, Y, P). By Proposition 15.6, any such P has a universal strongly P -generic condition. Hence by Corollary 3.10, P has the κ-covering property.

Finally, we prove that for most transitive models, the empty condition is a strongly generic condition.

Proposition 15.8. Suppose that X is an elementary substructure of (A, Y, P) of size κ such that X ∩ κ+ ∈ κ+ and X<κ ⊆ X. Then the pair (∅, ∅) is a strongly X-generic condition. Proof. Let θ := X ∩ κ+. Since X<κ ⊆ X, θ has cofinality κ. Let D be a dense subset of P ∩ X, and we will show that D is predense in P. Let p be a condition. Let M0,...,Mk−1 and P0,...,Pm−1 enumerate the members of Ap and Bp respectively. Note that since X<κ ⊆ X, for any model K on either of these lists, i ~ K ∩ θ ∈ X. For each i < k, let hQn : n < ωi enumerate the S-strong models in <κ i Mi ∩ Y. Since X ⊆ X, for each n < ω, Qn ∩ θ ∈ X. Therefore the sequence i hQn ∩ θ : n < ωi is in X. Note that the assumptions of the first and second paragraphs of Proposition 13.3 hold for A = Ap and B = Bp. i The objects p, θ, M0,...,Mk−1, P0,...,Pm−1, and hQn : n < ωi for i < k witness that (A, Y, P) satisfies the following statement:

0 0 0 0 0 i There exist v, θ , M0,...,Mk−1, P0,...,Pm−1, and hRn : n < ωi for i < k such that: (1) v ∈ P; 0 0 0 0 (2) M0,...,Mk−1 are in Av and P0,...,Pm−1 are in Bv; (3) cf(θ0) = κ; 0 0 0 0 (4) Mi ∩ θ = Mi ∩ θ and Pj ∩ θ = Pj ∩ θ for i < k and j < m; i ~ 0 (5) for all i < k, hRn : n < ωi enumerates the S-strong models in Mi ∩ Y, and i i 0 for all n < ω, Qn ∩ θ = Rn ∩ θ .

The parameters which appear in the above statement, namely κ, Mi ∩θ for i < k, i Pj ∩ θ for j < m, and hQn ∩ θ : n < ωi for i < k are all members of X. By the 0 0 0 0 0 i elementarity of X, we can fix v, θ , M0,...,Mk−1, P0,...,Pm−1, and hRn : n < ωi for i < k in X which satisfy the same statement. 0 0 For each M in Ap, let M denote Mi , where i < k and M = Mi. For each P in 0 0 Bp, let P denote Pj, where j < m and P = Pj. Since D is a dense subset of P ∩ X, we can fix w ≤ v in D. Let us show that w and p are compatible. This proves that D is predense in P, finishing the proof. It suffices to show that the pair

(Ap ∪ Aw,Bp ∪ Bw) MITCHELL’S THEOREM REVISITED 91 is a condition. Note that the assumptions of the third paragraph of Proposition 13.3 hold for C = Aw and D = Bw. So by Proposition 13.3, (Ap ∪ Aw,Bp ∪ Bw) is an S~-obedient side condition. Therefore this pair is in P, and it is obviously below p and w. Corollary 15.9. The forcing poset P is κ+-c.c. Proof. Since (∅, ∅) is the maximum element of P, by Proposition 3.11 it suﬃces to + show that there are stationarily many X in Pκ+ (H(κ )) for which (∅, ∅) is strongly X-generic. By Proposition 15.8, it suﬃces to show that there are stationarily many + + + <κ X in Pκ+ (H(κ )) such that X ∩ κ ∈ κ and X ⊆ X. But this follows easily <κ from the fact that κ = κ.

§16. The ﬁnal argument

We now complete the proof of Mitchell’s theorem. We begin by noting that the forcing poset P has the desired eﬀect on cardinal structure.

Proposition 16.1. The forcing poset P preserves ω1, collapses κ to become ω2, and is κ+-c.c. Proof. Immediate from Proposition 3.12, Lemma 15.2, and Corollaries 15.4, 15.7, and 15.9. Next we will show that we can apply the factorization theorem, Theorem 6.4. It is easy to see that P has greatest lower bounds. Namely, if (A, B) and (C,D) are in P and are compatible, then (A ∪ C,B ∪ D) is the greatest lower bound of (A, B) and (C,D).

Lemma 16.2. The forcing poset P satisfies property ∗(P, P). See Definition 6.2 for the definition of ∗.

Proof. Let p, q, and r be pairwise compatible conditions in P. Then q ∧ r = (Aq ∪ Ar,Bq ∪ Br), p ∧ q = (Ap ∪ Aq,Bp ∪ Bq), and p ∧ r = (Ap ∪ Ar,Bp ∪ Br). To see that p is compatible with q ∧ r, it suﬃces to show that

(Ap ∪ Aq ∪ Ar,Bp ∪ Bq ∪ Br) is an S~-obedient side condition. But looking over the requirements of being S~- obedient, any violation of these requirements involves an incompatibility between two objects appearing in the components of the pair, and hence would lead to a violation of the same requirement for one of the triples p ∧ q, p ∧ r, or q ∧ r. Proposition 16.3. Let Q ∈ Y be S~-strong such that cf(sup(Q)) = Q ∩ κ and Q ≺ (A, Y, P). Let qQ := (∅, {Q}). Let G be a generic filter on P which contains qQ. Then G ∩ Q is a V -generic filter on P ∩ Q, and V [G] = V [G ∩ Q][H], where H is a V [G∩Q]-generic filter on (P/qQ)/(G∩Q). Moreover, the pair (V [G∩Q],V [G]) satisfies the ω1-approximation property.

Proof. By Proposition 15.6, qQ is a universal strongly Q-generic condition. By + Propositions 7.20 and 15.3, there are stationarily many models in Pω1 (H(κ) ) which have universal strongly generic conditions. By Lemma 16.2, P satisﬁes property ∗(P, P). So the assumptions of Theorem 6.4 are satisﬁed, and we are done. 92 THOMAS GILTON AND JOHN KRUEGER

We will need the following technical lemma about names.

Lemma 16.4. Suppose that Q ∈ Y with Q ≺ (H(κ+), ∈, P), and q is a strongly Q-generic condition. Let G be a V -generic ﬁlter on P which contains q. Let a˙ ∈ Q be a nice P-name for a set of ordinals, and suppose that a˙ G is a subset of Q ∩ κ. Then a˙ G ∈ V [G ∩ Q].

Proof. Note that since q is strongly Q-generic, G ∩ Q is a V -generic filter on P ∩ Q by Lemma 3.3. Let α := Q∩κ. Sincea ˙ is a nice name, for each γ < α there is a unique antichain Aγ such that (p, γˇ) ∈ a˙ iff p ∈ Aγ . Sincea ˙ ∈ Q, by elementarity each Aγ is in Q. We claim that for all γ < α, G γ ∈ a˙ iff Aγ ∩ G ∩ Q 6= ∅. Sincea ˙ G ⊆ Q ∩ κ = α, it follows thata ˙ G is definable in V [G ∩ Q] from the sequence G hAγ : γ < αi and the set G ∩ Q. Thereforea ˙ ∈ V [G ∩ Q], which finishes the proof. If p ∈ Aγ ∩ G ∩ Q, then (p, γˇ) ∈ a˙ by the choice of Aγ . Since p ∈ G, it follows G G G that (ˇγ) = γ is ina ˙ . This shows that Aγ ∩ G ∩ Q 6= ∅ implies that γ ∈ a˙ . G Conversely, assume that γ ∈ a˙ . Then by the choice of Aγ , we can fix p ∈ G∩Aγ . So to show that Aγ ∩ G ∩ Q is nonempty, it suffices to show that p ∈ Q. Since Aγ ∈ Q is an antichain, by elementarity there is a maximal antichain A ∈ Q with Aγ ⊆ A. Let D be the dense set of u ∈ P such that for some s ∈ A, u ≤ s. By elementarity, D ∈ Q, and therefore D ∩ Q is dense in P ∩ Q by elementarity. Since q is strongly Q-generic, D ∩ Q is predense below q. As q ∈ G and D ∩ Q is predense below q, we can fix u ∈ G ∩ D ∩ Q. By elementarity and the definition of D, there is s ∈ A ∩ Q such that u ≤ s. Since u ∈ G, s ∈ G. Now p ∈ Aγ and Aγ ⊆ A, so p ∈ A. Also s ∈ A. Since s and p are both in G, they are compatible. But A is an antichain, so s = p. Since s ∈ Q, p ∈ Q. Proposition 16.5. Let τ < κ+ be an ordinal with cofinality κ which is closed under ∗ ˙ ˙ H . Let Y and Dτ be P-names such that P forces

Y˙ = {P : ∃p ∈ G˙ (P ∈ Bp)} and D˙ τ = {P ∩ κ : P ∈ Y˙ , τ ∈ P }. Suppose that β < κ is an ordinal with uncountable coﬁnality, and p is a condition which forces that β is a limit point of D˙ τ . Let Q := Sk(Aτ,β). Then: (1) Q ∈ Y; + (2) Q ∩ κ = β and Q ∩ κ = Aτ,β; (3) β ∈ Sτ+1; (4) cf(sup(Q)) = β; (5) Q is S~-strong; (6) p forces that Q is in Y˙ . Proof. Deﬁne Z∗ := {P ∈ Y : P is S~-strong,P ∩ κ < β, τ ∈ P }, and Z := {P ∩ τ : P ∈ Z∗}. ∗ Note that by Lemma 7.28, if P1 and P2 are in Z and P1 ∩ κ ≤ P2 ∩ κ, then P1 ∩ τ ⊆ P2 ∩ τ. MITCHELL’S THEOREM REVISITED 93

We claim that for all γ < β, there is P ∈ Z∗ such that γ < P ∩ κ. Namely, since p forces that β is a limit point of D˙ τ , there is q ≤ p and P ∈ Bq such that τ ∈ P ∗ and γ < P ∩ κ < β. Since P ∈ Bq, it follows that P ∈ Y is S~-strong. So P ∈ Z and γ < P ∩ κ, proving the claim. Consequently, (S Z) ∩ κ = β. S ∗ Next we claim that Z = Aτ,β. First, suppose that P ∈ Z , and we will show that P ∩ τ ⊆ Aτ,β. Since τ ∈ P , it follows that P ∩ τ = Aτ,P ∩κ by Lemma 7.27, S which is a subset of Aτ,β since P ∩κ < β. This shows that Z ⊆ Aτ,β. Conversely, S let ξ ∈ Aτ,β, and we will show that ξ ∈ Z. Since β is a limit ordinal, we can ∗ ﬁx γ < β such that ξ ∈ Aτ,γ . By the ﬁrst claim, there is P ∈ Z such that γ < P ∩ κ. Since γ < P ∩ κ and ξ ∈ Aτ,γ , it follows that ξ ∈ Aτ,P ∩κ. But τ ∈ P , so Aτ,P ∩κ = P ∩ τ by Lemma 7.27. Hence ξ ∈ P ∩ τ. As P ∩ τ ∈ Z, we have that ξ ∈ S Z. Since τ is closed under H∗, every set in Z is closed under H∗. As Z is a ⊆-chain, S ∗ + Z = Aτ,β is also closed under H . In particular, Q ∩ κ = Aτ,β. As noted above, [ Q ∩ κ = Aτ,β ∩ κ = ( Z) ∩ κ = β. This proves (2). ∗ By Lemma 7.28(2), if P1 and P2 are in Z and P1 ∩κ < P2 ∩κ, then sup(P1 ∩τ) ∈ S P2 ∩ τ, and hence sup(P1 ∩ τ) < sup(P2 ∩ τ). In particular, since Aτ,β = Z, ∗ sup(Aτ,β) = sup{sup(P ∩ τ): P ∈ Z }.

Since β has uncountable coﬁnality, sup(Aτ,β) = sup(Q) has uncountable coﬁnality. It follows that Q ∈ Y by Lemma 7.15.

Now we prove that Q ∩ κ = β is in Sτ+1. Fix M in X such that p, β, and τ are in M. Then q := (Ap ∪ {M},Bp) is in P and q ≤ p. Since β has uncountable coﬁnality, sup(M ∩ β) < β. As q forces that β is a limit point of D˙ τ , we can ﬁx r ≤ q and P ∈ Br such that sup(M ∩ β) < P ∩ κ < β and τ ∈ P . It follows that β = min((M∩κ)\(P ∩κ)). As τ ∈ M∩P , τ+1 ∈ M∩P by elementarity. So M ∈ Ar, + P ∈ Br, and τ + 1 ∈ M ∩ P ∩ κ , which implies that β = min((M ∩ κ) \ (P ∩ κ)) is in Sτ+1 by the fact that (Ar,Br) is S~-obedient.

Since β ∈ Sτ+1, β is inaccessible, and in particular is regular. Therefore the ∗ ordinal sup(Aτ,β) = sup{sup(P ∩ τ): P ∈ Z } is the supremum of a sequence of ordinals of order type β. It follows that sup(Q) has coﬁnality β. So cf(sup(Q)) = Q ∩ κ. Now we show that Q is S~-strong. Since β ∈ Sτ+1, β ∈ Sτ . As τ is a limit + ordinal, for all ξ ∈ Aτ,β, β ∈ Sξ. So if ξ ∈ Q ∩ κ = Aτ,β, then Q ∩ κ = β ∈ Sξ, which shows that Q is S~-strong.

It remains to show that p forces that Q is in Y˙ . It suffices to prove that for all q ≤ p, there is r ≤ q such that Q ∈ Br. So let q ≤ p. We claim that (Aq,Bq ∪ {Q}) is a condition below q. Since Q is S~-strong, to prove that (Aq,Bq ∪{Q}) is an S~-obedient side condition, it suffices to show that if M ∈ Aq and ζ = min((M ∩ κ) \ β), then for all σ ∈ + M ∩ Q ∩ κ , ζ ∈ Sσ. + + Let σ ∈ M ∩ Q ∩ κ . Since σ ∈ Q ∩ κ = Aτ,β and β is a limit ordinal, we can fix γ < β such that σ ∈ Aτ,γ . By increasing γ if necessary, also assume that 94 THOMAS GILTON AND JOHN KRUEGER sup(M ∩ β) < γ. As q forces that β is a limit point of D˙ τ , we can fix s ≤ q and P ∈ Bs such that γ < P ∩ κ < β and τ ∈ P . So

Aτ,γ ⊆ Aτ,P ∩κ = P ∩ τ. In particular, σ ∈ P . Since ζ = min((M ∩ κ) \ β), we have that sup(M ∩ ζ) = sup(M ∩ β) < γ < P ∩ κ < β ≤ ζ.

+ So M ∈ As, P ∈ Bs, σ ∈ M ∩ P ∩ κ , and ζ = min((M ∩ κ) \ (P ∩ κ)). Therefore ~ ζ ∈ Sσ since (As,Bs) is an S-obedient side condition.

+ Lemma 16.6. Let τ < κ be an ordinal of coﬁnality κ, and let β ∈ Sτ+1. Suppose + that Q ∈ Y is S~-strong, Q ∩ κ = β, Q ∩ κ = Aτ,β, and cf(sup(Q)) = β. Then the set {P ∈ Q ∩ Y : P is S~-strong, cf(sup(P )) = P ∩ κ} is stationary in Pβ(Q). Proof. Let F : Q<ω → Q, and we will ﬁnd P ∈ Q∩Y such that cf(sup(P )) = P ∩κ, + P is S~-strong, and P is closed under F . Since Q ∈ Y, Q ∩ κ = Aτ,β is closed ∗ + under H . As Q ∩ κ is the union of the increasing and continuous chain {Aτ,i : i < β}, there exists a club C ⊆ β such that for all α ∈ C, Aτ,α is closed under ∗ H and Aτ,α ∩ κ = α. Then Q is the union of the increasing and continuous chain {Sk(Aτ,α): α ∈ C}. Fix a club D ⊆ C such that for all α ∈ D, Sk(Aτ,α) is closed under F . For each α ∈ D, let Qα := Sk(Aτ,α). We claim that for all α ∈ D, Qα ∈ Q. + + Since |Qα ∩ κ | = |Aτ,α| < |α| < β and cf(sup(Q)) = β, it follows that Aτ,α is a + bounded subset of Q ∩ κ = Aτ,β. Also

cf(sup(Qα)) ≤ α < β = cf(Q ∩ κ).

By Lemma 7.14, sup(Qα) ∈ Q. But Aτ,α = Asup(Aτ,α),α by coherence, and since sup(Aτ,α) and α are in Q, so is Aτ,α by elementarity. Hence Qα ∈ Q by elementarity. Fix a club E ⊆ lim(D) such that for all α ∈ E, for all γ ∈ α ∩ D, Qγ ∈ Qα. S In particular, for all α ∈ E, since Qα = {Qγ : γ ∈ α ∩ D}, it follows that cf(sup(Qα)) = cf(ot(α ∩ D)). So if α ∈ E is regular, then cf(sup(Qα)) = α. Since β ∈ Sτ+1, Sτ ∩ β is stationary in β. So we can fix α ∈ E ∩ Sτ . To finish the proof, it suffices to show that Qα = Sk(Aτ,α) is in Q ∩ Y, Qα is S~-strong, cf(sup(Qα)) = Qα ∩ κ = α, and Qα is closed under F . We know that Qα is closed under F by the definition of D. We previously observed that Qα ∈ Q, and since α ∈ E is regular, cf(sup(Qα)) = α. In particular, + Qα ∈ Y by Lemma 7.15. To see that Qα is S~-strong, let ξ ∈ Qα ∩ κ = Aτ,α. Since α ∈ Sτ and τ is a limit ordinal, α ∈ Sη for all η ∈ Aτ,α. In particular, Qα ∩ κ = α ∈ Sξ.

Lemma 16.7. Suppose that Q ∈ Y is S~-strong and Q ≺ (A, Y, P). Let β := Q ∩ κ. Suppose that the set {P ∈ Q ∩ Y : P is S~-strong, cf(sup(P )) = P ∩ κ} is stationary in Pβ(Q). Then the forcing poset P ∩ Q forces that β is a regular cardinal. MITCHELL’S THEOREM REVISITED 95

Proof. Let γ < β, and let f˙ be a (P ∩ Q)-name for a function from γ to β. Fix a condition p ∈ P ∩ Q, and we will find q ≤ p in P ∩ Q which forces that f˙ is bounded ˙ in β. Let F be the set of triples (u, i, ξ) such that u ∈ P ∩ Q and u P∩Q f(i) = ξ. Let {gn : n < ω} be a set of definable Skolem functions for the structure (A, Y, P). Since Q ≺ (A, Y, P), Q is closed under gn for all n < ω. By the assumption of the lemma, we can fix P ∈ Q ∩ Y such that P is S~-strong, cf(sup(P )) = P ∩ κ, P is closed under gn for all n < ω, and P ≺ (Q, ∈, P ∩ Q, p, γ, F ). In particular, P ≺ (A, Y, P). As p ∈ P , Proposition 15.6 implies that q := (Ap,Bp ∪ {P }) is a strongly P -generic condition below p. We claim that ˙ q P∩Q ran(f) ⊆ P ∩ κ. Since P ∩ κ < Q ∩ κ = β, this completes the proof. Let i < γ, and we will show that ˙ q P∩Q f(i) ∈ P ∩ κ. Let D be the set of s ∈ P ∩ P such that for some ξ ∈ P ∩ κ,(s, i, ξ) ∈ F . We claim that D is dense in P ∩ P . So let u ∈ P ∩ P be given. Then u ∈ P ∩ Q. Since f˙ is a (P ∩ Q)-name for a function from γ to β, there is v ≤ u and ξ < Q ∩ κ such ˙ that v P∩Q f(i) = ξ, and hence (v, i, ξ) ∈ F . Since P ≺ (Q, ∈, P ∩ Q, p, γ, F ) and u and i are in P , by elementarity there is v ∈ P and ξ ∈ P ∩ κ such that v ≤ u and (v, i, ξ) ∈ F . Then v ≤ u and v ∈ D. Since q is strongly P -generic, D is predense in P below q. Let r ≤ q in Q ∩ P decide the value of f˙(i) to be ξ, and we will show that ξ ∈ P . Then r ≤ q is in P, and (r, i, ξ) ∈ F . Since D is predense in P below q, for some u ∈ D, r and u are compatible in P. By the elementarity of Q, P ∩ Q is closed under greatest lower bounds, and therefore r and u are also compatible in P ∩ Q. Since u ∈ D, by the definition of D there is ξ0 ∈ P ∩ κ such that (u, i, ξ0) ∈ F . But (r, i, ξ) ∈ F and (u, i, ξ0) ∈ F imply, by the compatibility of r and u in P ∩ Q, that ξ = ξ0. Since 0 ξ ∈ P , it follows that ξ ∈ P .

Recall that for a sequence ~a = hai : i < ω2i of countable sets, S~a is the set of limit ordinals α < ω2 for which there exists a club c ⊆ α with order type cf(α) such that for all β < α, there is i < α with c ∩ β = ai. A set S is in the approachability ideal I[ω2] iﬀ there exists such a sequence ~a and a club D with S ∩ D ⊆ S~a. In particular, if I[ω2] contains a stationary subset of ω2 ∩ cof(ω1), then for some sequence ~a, S~a ∩ cof(ω1) is stationary. We will show that this last statement fails in any generic extension by P. Theorem 16.8. The forcing poset P forces that there is no stationary subset of ω2 ∩ cof(ω1) in the approachability ideal I[ω2].

Proof. Suppose for a contradiction that p is a condition, ~a = ha˙ i : i < κi is a ˙ sequence of P-names for countable subsets of κ, and p forces that S~a ∩ cof(ω1) is stationary. Without loss of generality, assume that eacha ˙ i is a nice name, which i means that for some sequence of antichains hAα : α < κi of P,a ˙ i is equal to the i + set of pairs {(p, αˇ): p ∈ Aα, α < κ}. As P is κ -c.c., each namea ˙ i is a member of + + ∗ ∗ + + H(κ ). Fix γ < κ such that for all i < κ,a ˙ i is in f [γ], where f : κ → H(κ ) is the bijection described in Notation 7.1. Let M be an elementary substructure of (A, Y, P) such that |M| = κ, M ∩ κ+ ∈ κ+ ∩ cof(κ), and γ < M ∩ κ+. Let τ := M ∩ κ+. Since γ < M ∩ κ+ and M is closed 96 THOMAS GILTON AND JOHN KRUEGER

∗ ˙ ˙ under f , it follows that for all i < κ,a ˙ i is in M. Fix P-names Y and Dτ such that P forces

Y˙ = {P : ∃p ∈ G˙ (P ∈ Bp)} and D˙ τ = {P ∩ κ : P ∈ Y˙ , τ ∈ P }.

An easy observation which follows from Lemma 15.5 is that D˙ τ is forced to be cofinal in κ. Therefore lim(D˙ τ ) is forced to be club in κ. For each α < κ, let Qα := Sk(Aτ,α). Since τ is the union of the increasing and continuous sequence {Aτ,α : α < κ}, clearly M = Sk(τ) is the union of the increasing and continuous sequence {Qα : α < κ}. Let E be a club subset of κ such + that for all α ∈ E, Qα ∩ κ = α, Qα ∩ κ = Aτ,α, Qα ≺ (A, Y, P), and for all i < α, a˙ i ∈ Qα. ˙ ˙ Clearly p forces that E ∩ lim(Dτ ) is club in κ. Since p forces that S~a ∩ cof(ω1) is stationary, we can fix q ≤ p and α < κ such that q forces that α is in E ∩ lim(D˙ τ ) ∩ ˙ S~a ∩ cof(ω1). Since q forces that cf(α) = ω1, clearly α has uncountable cofinality. Let Q := Qα. Then by Proposition 16.5, Q ∈ Y is S~-strong, Q ∩ κ = α ∈ Sτ+1, + Q ∩ κ = Aτ,α, cf(sup(Q)) = Q ∩ κ = α, and q forces that Q is in Y˙ . By extending q if necessary, we can assume without loss of generality that Q ∈ Bq. By Lemma 16.6, the set {P ∈ Q ∩ Y : P is S~-strong, cf(sup(P )) = P ∩ κ} is stationary in Pα(Q). By Lemma 16.7, the forcing poset P ∩ Q forces that α is a regular cardinal. Since Q ∈ Bq, clearly q ≤ qQ := (∅, {Q}). Let G be a V -generic filter on P containing q, and we will get a contradiction by considering the generic extension V [G]. Since q ≤ qQ, it follows that qQ ∈ G. By Proposition 16.3, V [G] can be factored as V [G] = V [G ∩ Q][H], where G ∩ Q is a V -generic filter on P ∩ Q, H is a V [G ∩ Q]-generic filter on (P/qQ)/(G ∩ Q), and the pair (V [G ∩ Q],V [G]) has the ω1-approximation property. As α is in S~a ∩ cof(ω1), in V [G] there is a club c ⊆ α with order type ω1 such G that for all β < α, there is i < α such that c ∩ β =a ˙ i . For any such β and i, G a˙ i ∈ Q, anda ˙ i = c ∩ β is a subset of Q ∩ κ = α. By Lemma 16.4, it follows that G a˙ i ∈ V [G ∩ Q]. So for all β < α, c ∩ β ∈ V [G ∩ Q]. By Lemma 6.1, c ∈ V [G ∩ Q]. But since c has order type ω1, it follows that α has cofinality ω1 in V [G ∩ Q]. Now α ∈ Sτ+1, and in particular, α is inaccessible in V , but α is not regular in V [G ∩ Q]. However, we previously observed that P ∩ Q forces that α is a regular cardinal, so we have a contradiction. References [1] J. Baumgartner, A. Taylor, and S. Wagon. On splitting stationary subsets of large cardinals. J. Symbolic Logic, 42(2):203–214, 1977. [2] S. Cox and J. Krueger. Quotients of strongly proper forcings and guessing models. J. Symbolic Logic, 81(1):264–283, 2016. [3] S.D. Friedman. Forcing with finite conditions. In Set Theory: Centre de Recerca Matemàtica, Barcelona, 2003-2004, Trends in Mathematics, pages 285–295. BirkhäuserVerlag, 2006. [4] S.D. Friedman and J. Krueger. Thin stationary sets and disjoint club sequences. Trans. Amer. Math. Soc., 359:2407–2420, 2007. [5] J. Hamkins. Extensions with the approximation and cover properties have no new large cardinals. Fund. Math., 180:257–277, 2003. [6] J. Krueger. Forcing with adequate sets of models as side conditions. To appear in Mathemat- ical Logic Quarterly. MITCHELL’S THEOREM REVISITED 97

[7] J. Krueger. Coherent adequate sets and forcing square. Fund. Math., 224:279–300, 2014. [8] J. Krueger. Strongly adequate sets and adding a club with finite conditions. Arch. Math. Logic, 53(1-2):119–136, 2014. [9] J. Krueger. Adding a club with finite conditions, part II. Arch. Math. Logic, 54(1-2):161–172, 2015. [10] J. Krueger and M.A. Mota. Coherent adequate forcing and preserving CH. J. Math. Log., 15(2), 2015. [11] W. Mitchell. Aronszajn trees and the independence of the transfer property. Ann. Math. Logic, pages 21–46, 1972. [12] W. Mitchell. I[ω2] can be the nonstationary ideal on Cof(ω1). Trans. Amer. Math. Soc., 361(2):561–601, 2009. [13] I. Neeman. Forcing with sequences of models of two types. Notre Dame J. Form. Log., 55(2):265–298, 2014. [14] S. Shelah. Reflecting stationary sets and successors of singular cardinals. Arch. Math. Logic, 31(1):25–53, 1991. [15] S. Todorˇcević.A note on the proper forcing axiom. In Axiomatic set theory (Boulder, Colo., 1983), volume 31 of Contemp. Math., pages 209–218. Amer. Math. Soc., Providence, RI, 1984.

Thomas Gilton, Department of Mathematics, University of California, Los Angeles, Box 951555, Los Angeles, CA 90095-1555 E-mail address: [email protected]

John Krueger, Department of Mathematics, University of North Texas, 1155 Union Circle #311430, Denton, TX 76203 E-mail address: [email protected]